Applications of Derivatives (Tangent Lines, Optimization)

Derivatives play a crucial role in many real-world applications, including finding the equations of tangent lines and solving optimization problems. Derivatives provide us with rates of change, slopes of curves, and help us determine the maximum and minimum values of functions, which is key in optimization.

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1. Tangent Lines

One of the most fundamental applications of derivatives is finding the equation of the tangent line to a curve at a given point. The derivative of a function at a particular point gives the slope of the tangent line to the curve at that point.

The equation of a tangent line to the curve $y=f(x)$ at the point $(x_1,y_1)$ is given by:

$y–y_1=m(x–x_1)$

Where $m$ is the slope of the tangent line, which is the value of the derivative of $f(x)$ at $x=x_1$, that is:

$m=f^{\prime }(x_1)$

Example 1: Finding the Equation of a Tangent Line

Problem:
Find the equation of the tangent line to the curve $y=x^2+2x+1$ at the point where $x=1$.

Answer:

Step 1: Given Data:
The function is $y=x^2+2x+1$, and we need the tangent line at $x=1$.

Step 2: Solution:
First, find the derivative of the function to get the slope of the tangent line:
$f(x)=x^2+2x+1$
$f^{\prime }(x)=2x+2$

Now, evaluate the derivative at $x=1$:
$f^{\prime }(1)=2(1)+2$
$=4$

So, the slope of the tangent line is $m=4$.

Next, find the coordinates of the point of tangency:
$y_1=f(1)=(1{)}^2+2(1)+1$
$=1+2+1$
$=4$

The point of tangency is $(1,4)$.

Now, write the equation of the tangent line using the point-slope form:
$y–y_1=m(x–x_1)$
$y–4=4(x–1)$

Step 3: Final Answer:
$y=4x$

Thus, the equation of the tangent line is $y=4x$.

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2. Optimization

In calculus, optimization involves finding the maximum or minimum values of a function. These can represent the highest or lowest points on a graph, the largest or smallest values a quantity can take, or even the most efficient way to do something in real life. The derivative is key in solving these types of problems because the points where the derivative equals zero (called critical points) can be potential maxima or minima.

To solve an optimization problem:

1. Find the derivative of the function.
2. Set the derivative equal to zero to find critical points.
3. Test the critical points (and endpoints, if applicable) to determine whether they are maxima, minima, or neither.

Example 2: Optimization Problem

Problem:
A farmer wants to build a rectangular fence using 100 meters of fencing. What dimensions should the farmer use to maximize the area of the fenced region?

Answer:

Step 1: Given Data:
The perimeter of the rectangle is 100 meters, so:
$2l+2w=100$, where $l$ is the length and $w$ is the width.

We need to maximize the area, $A=l\times w$.

Step 2: Solution:
From the perimeter equation, solve for $l$:
$l=\frac{100–2w}{2}$
$=50–w$

Now substitute this into the area formula:
$A=l\times w$
$A=(50–w)w$
$A=50w–w^2$

To find the maximum area, take the derivative of $A$ with respect to $w$:
$\frac{dA}{dw}=50–2w$

Set the derivative equal to zero to find the critical points:
$50–2w=0$
$2w=50$
$w=25$

Now substitute $w=25$ into the perimeter equation to find $l$:
$l=50–w$
$l=50–25$
$l=25$

Step 3: Final Answer:
The dimensions of the rectangle that maximize the area are $l=25$ meters and $w=25$ meters, so the maximum area occurs when the fence forms a square.

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3. Maximizing Revenue or Profit

Another common application of derivatives is in economics, particularly when trying to maximize revenue or profit.

Example 3: Maximizing Profit

Problem:
A company’s profit function is given by $P(x)=-5x^2+400x–1500$, where $x$ is the number of products sold. How many products should the company sell to maximize its profit?

Answer:

Step 1: Given Data:
The profit function is $P(x)=-5x^2+400x–1500$.

Step 2: Solution:
First, find the derivative of the profit function:
$\frac{dP}{dx}=-10x+400$

Now, set the derivative equal to zero to find the critical points:
$-10x+400=0$
$10x=400$
$x=40$

To determine if this is a maximum, take the second derivative:
$\frac{d^{2}P}{d{x}^2}=-10$

Since the second derivative is negative, the function has a maximum at $x=40$.

Step 3: Final Answer:
The company should sell 40 products to maximize its profit.

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4. Minimizing Cost

In some cases, derivatives are used to minimize costs in manufacturing or construction problems.

Example 4: Minimizing Cost

Problem:
A company needs to design a cylindrical can with a volume of 1000 cubic centimeters. What dimensions of the can will minimize the cost of materials if the cost is proportional to the surface area?

Answer:

Step 1: Given Data:
The volume of the cylinder is $V=\pi r^{2}h=1000$ cm${}^3$, where $r$ is the radius and $h$ is the height.
The surface area of the can is $A=2\pi r^2+2\pi rh$.

Step 2: Solution:
First, solve the volume equation for $h$:
$h=\frac{1000}{\pi r^2}$

Now substitute this into the surface area equation:
$A=2\pi r^2+2\pi r\left(\frac{1000}{\pi r^2}\right)$
$A=2\pi r^2+\frac{2000}{r}$

To minimize the surface area, take the derivative of $A$ with respect to $r$:
$\frac{dA}{dr}=4\pi r–\frac{2000}{r^2}$

Set the derivative equal to zero to find the critical points:
$4\pi r=\frac{2000}{r^2}$
$4\pi r^3=2000$
$r^3=\frac{2000}{4\pi }$
$r=\sqrt[3]{\frac{500}{\pi }}$

Step 3: Final Answer:
The radius that minimizes the cost is approximately $r\approx 5.42$ cm. You can substitute this value back into the volume equation to find the height.

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Conclusion

The tangent line to a curve gives the slope and the equation of the line that just touches the curve at a specific point, which is useful for approximating values of functions. In optimization problems, derivatives help us find the maximum and minimum values of functions, whether we are trying to maximize profits, minimize costs, or optimize the use of resources.

By understanding the concepts of critical points, maxima, and minima, you can tackle a wide variety of real-world problems where optimization plays a critical role.

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Example 1: Finding the Equation of a Tangent Line

Problem:
Find the equation of the tangent line to the curve $y=x^3–3x^2+2$ at the point where $x=1$.

Answer:
Step 1: Given Data:
The function is $y=x^3–3x^2+2$. We need the tangent line at $x=1$.

Step 2: Solution:
First, find the derivative of the function to get the slope of the tangent line:
$f^{\prime }(x)=3x^2–6x$

Now, evaluate the derivative at $x=1$:
$f^{\prime }(1)=3(1{)}^2–6(1)$
$=3–6$
$=-3$

So, the slope of the tangent line is $m=-3$.

Next, find the coordinates of the point of tangency:
$y_1=f(1)=(1{)}^3–3(1{)}^2+2$
$=1–3+2$
$=0$

The point of tangency is $(1,0)$.

Now, write the equation of the tangent line using the point-slope form:
$y–y_1=m(x–x_1)$
$y–0=-3(x–1)$

Step 3: Final Answer:
$y=-3x+3$

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Example 2: Optimization Problem (Maximizing Area)

Problem:
A rectangular garden is to be fenced using 60 meters of fencing. What dimensions should the garden have to maximize the area?

Answer:
Step 1: Given Data:
Let the length be $l$ meters and the width be $w$ meters. The perimeter is given by:
$2l+2w=60$

We need to maximize the area:
$A=l\times w$

Step 2: Solution:
From the perimeter equation, solve for $l$:
$l=30–w$

Substituting into the area formula gives:
$A=(30–w)w$
$A=30w–w^2$

To find the maximum area, take the derivative of $A$ with respect to $w$:
$\frac{dA}{dw}=30–2w$

Set the derivative equal to zero to find critical points:
$30–2w=0$
$2w=30$
$w=15$

Now, substitute $w=15$ back into the perimeter equation to find $l$:
$l=30–15$
$l=15$

Step 3: Final Answer:
The dimensions that maximize the area are $l=15$ meters and $w=15$ meters (a square).

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Example 3: Maximizing Profit

Problem:
A company’s profit function is given by $P(x)=-2x^2+80x–200$, where $x$ is the number of items sold. How many items should the company sell to maximize profit?

Answer:
Step 1: Given Data:
The profit function is $P(x)=-2x^2+80x–200$.

Step 2: Solution:
First, find the derivative of the profit function:
$\frac{dP}{dx}=-4x+80$

Set the derivative equal to zero to find critical points:
$-4x+80=0$
$4x=80$
$x=20$

To determine if this is a maximum, take the second derivative:
$\frac{d^{2}P}{d{x}^2}=-4$

Since the second derivative is negative, the function has a maximum at $x=20$.

Step 3: Final Answer:
The company should sell 20 items to maximize profit.

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Example 4: Minimizing Cost

Problem:
A manufacturer produces a product at a cost given by the function $C(x)=5x^2+20x+100$. What production level minimizes the cost?

Answer:
Step 1: Given Data:
The cost function is $C(x)=5x^2+20x+100$.

Step 2: Solution:
First, find the derivative of the cost function:
$\frac{dC}{dx}=10x+20$

Set the derivative equal to zero to find critical points:
$10x+20=0$
$10x=-20$
$x=-2$

Since $x=-2$ does not make sense in the context of production, we evaluate the endpoints or check for a minimum in the positive domain.

Step 3: Final Answer:
The minimum cost does not occur at a feasible production level based on the critical point analysis. Therefore, we need to analyze feasible production levels.

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Example 5: Tangent Line at a Trigonometric Function

Problem:
Find the equation of the tangent line to the curve $y=\sin (x)$ at the point where $x=\frac{\pi }{4}$.

Answer:
Step 1: Given Data:
The function is $y=\sin (x)$, and we need the tangent line at $x=\frac{\pi }{4}$.

Step 2: Solution:
First, find the derivative of the function to get the slope of the tangent line:
$f^{\prime }(x)=\cos (x)$

Now, evaluate the derivative at $x=\frac{\pi }{4}$:
$f^{\prime }\left(\frac{\pi }{4}\right)=\cos \left(\frac{\pi }{4}\right)$
$=\frac{\sqrt{2}}{2}$

So, the slope of the tangent line is $m=\frac{\sqrt{2}}{2}$.

Next, find the coordinates of the point of tangency:
$y_1=f\left(\frac{\pi }{4}\right)=\sin \left(\frac{\pi }{4}\right)$
$=\frac{\sqrt{2}}{2}$

The point of tangency is $(\frac{\pi }{4},\frac{\sqrt{2}}{2})$.

Now, write the equation of the tangent line using the point-slope form:
$y–y_1=m(x–x_1)$
$y–\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(x–\frac{\pi }{4}\right)$

Step 3: Final Answer:
$y=\frac{\sqrt{2}}{2}x+\left(\frac{\sqrt{2}}{2}–\frac{\sqrt{2}}{8}\right)$

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Example 6: Optimization with a Fixed Cost

Problem:
A company can produce a fixed volume of 5000 cubic meters of a product using a rectangular box. What dimensions will minimize the surface area of the box if the height is fixed at 10 meters?

Answer:
Step 1: Given Data:
Let the length be $l$ meters, the width be $w$ meters, and the height is fixed at $h=10$ meters.

The volume of the box is given by:
$V=l\times w\times h=5000$

Thus, we have:
$lw=500$

The surface area of the box is given by:
$A=2lw+2lh+2wh$
$A=2lw+20l+20w$

Step 2: Solution:
Substituting $w$ from the volume equation into the surface area equation:
$w=\frac{500}{l}$
$A=2l\left(\frac{500}{l}\right)+20l+20\left(\frac{500}{l}\right)$
$A=1000+20l+\frac{10000}{l}$

To find the minimum area, take the derivative of $A$ with respect to $l$:
$\frac{dA}{dl}=20–\frac{10000}{l^2}$

Set the derivative equal to zero:
$20–\frac{10000}{l^2}=0$
$\frac{10000}{l^2}=20$
$10000=20l^2$
$l^2=500$
$l=\sqrt{500}\approx 22.36\text{ meters}$

Now find $w$:
$w=\frac{500}{\sqrt{500}}=\sqrt{500}\approx 22.36\text{ meters}$

Step 3: Final Answer:
The dimensions that minimize the surface area are approximately $l\approx 22.36$ meters and $w\approx 22.36$ meters with $h=10$ meters.

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Example 7: Maximizing the Area of a Triangle

Problem:
A triangle has a fixed base of 10 meters. What height should be used to maximize the area of the triangle?

Answer:
Step 1: Given Data:
The area of a triangle is given by:
$A=\frac{1}{2}\times \text{base}\times \text{height}$
Let the base be $b=10$ meters and height be $h$.

Step 2: Solution:
The area becomes:
$A=\frac{1}{2}\times 10\times h$
$A=5h$

To maximize the area, we can see that as $h$ increases, the area increases indefinitely. Thus, we need to ensure that height is within practical limits (such as the physical constraints of the triangle).

Step 3: Final Answer:
The area of the triangle increases with height; there is no maximum height constraint given in this problem, so the area can be made arbitrarily large.

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Example 8: Minimizing the Length of a Fence

Problem:
A farmer wants to fence a rectangular area of 2000 square meters. What dimensions should the farmer use to minimize the length of the fence?

Answer:
Step 1: Given Data:
Let the length be $l$ meters and the width be $w$ meters. The area is given by:
$A=l\times w=2000$

Step 2: Solution:
From the area equation, solve for $w$:
$w=\frac{2000}{l}$

The perimeter (length of the fence) is given by:
$P=2l+2w$
$P=2l+2\left(\frac{2000}{l}\right)$
$P=2l+\frac{4000}{l}$

To minimize the perimeter, take the derivative of $P$ with respect to $l$:
$\frac{dP}{dl}=2–\frac{4000}{l^2}$

Set the derivative equal to zero:
$2–\frac{4000}{l^2}=0$
$\frac{4000}{l^2}=2$
$4000=2l^2$
$l^2=2000$
$l=\sqrt{2000}\approx 44.72\text{ meters}$

Now find $w$:
$w=\frac{2000}{\sqrt{2000}}=\sqrt{2000}\approx 44.72\text{ meters}$

Step 3: Final Answer:
The dimensions that minimize the length of the fence are approximately $l\approx 44.72$ meters and $w\approx 44.72$ meters.

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Example 9: Maximizing Volume of a Box

Problem:
A box with a square base is to be made with a surface area of 600 square meters. What dimensions will maximize the volume of the box?

Answer:
Step 1: Given Data:
Let the side length of the base be $x$ meters and the height be $h$ meters. The surface area of the box is given by:
$A=x^2+4xh=600$

Step 2: Solution:
From the surface area equation, solve for $h$:
$4xh=600–x^2$
$h=\frac{600–x^2}{4x}$

Now, the volume of the box is given by:
$V=x^{2}h$
Substituting for $h$ gives:
$V=x^2\cdot \frac{600–x^2}{4x}$
$V=\frac{600x–x^3}{4}$

To find the maximum volume, take the derivative of $V$ with respect to $x$:
$\frac{dV}{dx}=\frac{600–3x^2}{4}$

Set the derivative equal to zero:
$600–3x^2=0$
$3x^2=600$
$x^2=200$
$x=\sqrt{200}\approx 14.14\text{ meters}$

Now find $h$:
$h=\frac{600–(14.14{)}^2}{4\cdot 14.14}$
$h=\frac{600–200}{56.56}\approx 7.07\text{ meters}$

Step 3: Final Answer:
The dimensions that maximize the volume are approximately $x\approx 14.14$ meters (base side) and $h\approx 7.07$ meters (height).

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Example 10: Minimizing Distance

Problem:
A point is located at $(3,4)$. Find the point on the line $y=2x+1$ that is closest to this point.

Answer:
Step 1: Given Data:
Let the point on the line be $(x,y)$, where $y=2x+1$. We need to minimize the distance $D$ from $(x,y)$ to $(3,4)$.

Step 2: Solution:
The distance formula is given by:
$D=\sqrt{(x–3{)}^2+(y–4{)}^2}$

Substituting for $y$ gives:
$D=\sqrt{(x–3{)}^2+(2x+1–4{)}^2}$
$=\sqrt{(x–3{)}^2+(2x–3{)}^2}$

To minimize $D$, we can minimize $D^2$:
$D^2=(x–3{)}^2+(2x–3{)}^2$
$=(x–3{)}^2+(4x^2–12x+9)$
$=5x^2–18x+18$

To find the minimum distance, take the derivative of $D^2$ with respect to $x$:
$\frac{d(D^2)}{dx}=10x–18$

Set the derivative equal to zero:
$10x–18=0$
$10x=18$
$x=1.8$

Now find $y$:
$y=2(1.8)+1=3.6+1=4.6$

Step 3: Final Answer:
The point on the line $y=2x+1$ closest to $(3,4)$ is approximately $(1.8,4.6)$.

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Example 11: Maximizing Area of a Triangle (Again)

Problem:
Given a fixed perimeter of a triangle of 60 meters, what dimensions yield the maximum area?

Answer:
Step 1: Given Data:
Let the lengths of the sides be $a$, $b$, and $c$. The perimeter is given by:
$a+b+c=60$

Step 2: Solution:
To maximize the area, we can use Heron’s formula:
$A=\sqrt{s(s-a)(s-b)(s-c)}$
where $s=\frac{a+b+c}{2}=30$.

Assuming $a=b=c=20$ (equilateral triangle):
$A=\sqrt{30(30-20)(30-20)(30-20)}$
$=\sqrt{30\cdot 10\cdot 10\cdot 10}$
$=\sqrt{3000}\approx 54.77{\text{ m}}^2$.

Step 3: Final Answer:
The dimensions of the triangle yielding the maximum area are each approximately 20 meters.

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Example 12: Cost Minimization Problem

Problem:
A company finds that the cost $C$ of producing $x$ items is given by $C(x)=2x^2+50x+300$. Find the production level that minimizes cost.

Answer:
Step 1: Given Data:
The cost function is $C(x)=2x^2+50x+300$.

Step 2: Solution:
First, find the derivative of the cost function:
$\frac{dC}{dx}=4x+50$

Set the derivative equal to zero to find critical points:
$4x+50=0$
$4x=-50$
$x=-12.5$

Since negative production doesn’t make sense, we evaluate the cost function at feasible production levels.

Step 3: Final Answer:
The production level cannot be determined from negative values, hence look for minimum costs at feasible limits.

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Example 13: Maximizing the Volume of a Rectangular Box

Problem:
A rectangular box with a square base is to have a volume of 500 cubic centimeters. What dimensions will maximize the surface area?

Answer:
Step 1: Given Data:
Let the side length of the base be $x$ cm and the height be $h$ cm. The volume of the box is given by:
$V=x^{2}h=500$

Step 2: Solution:
From the volume equation, solve for $h$:
$h=\frac{500}{x^2}$

The surface area $S$ of the box is given by:
$S=x^2+4xh$

Substituting for $h$ gives:
$S=x^2+4x\left(\frac{500}{x^2}\right)$
$S=x^2+\frac{2000}{x}$

To maximize the surface area, take the derivative of $S$ with respect to $x$:
$\frac{dS}{dx}=2x–\frac{2000}{x^2}$

Set the derivative equal to zero:
$2x–\frac{2000}{x^2}=0$
$2x^3=2000$
$x^3=1000$
$x=10\text{ cm}$

Now substitute back to find $h$:
$h=\frac{500}{{10}^2}=\frac{500}{100}=5\text{ cm}$

Step 3: Final Answer:
The dimensions that maximize the surface area are $x=10$ cm and $h=5$ cm.

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Example 14: Profit Maximization for a Business

Problem:
A company’s profit function is given by $P(x)=-2x^2+80x–300$, where $x$ is the number of items sold. How many items should be sold to maximize profit?

Answer:
Step 1: Given Data:
The profit function is $P(x)=-2x^2+80x–300$.

Step 2: Solution:
First, find the derivative of the profit function:
$\frac{dP}{dx}=-4x+80$

Set the derivative equal to zero to find critical points:
$-4x+80=0$
$4x=80$
$x=20$

To determine if this is a maximum, take the second derivative:
$\frac{d^{2}P}{d{x}^2}=-4$

Since the second derivative is negative, the function has a maximum at $x=20$.

Step 3: Final Answer:
The company should sell 20 items to maximize profit.

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Example 15: Minimizing Travel Time

Problem:
A car travels from point A to point B, a distance of 120 km. The car’s speed varies according to the function $v(t)=60–0.5t$, where $t$ is the time in hours. Find the time at which the car should travel to minimize the time taken.

Answer:
Step 1: Given Data:
The distance is $d=120$ km, and speed is given by $v(t)=60–0.5t$.

Step 2: Solution:
We know that time is given by:
$t=\frac{d}{v}$
Substituting for $v(t)$ gives:
$t=\frac{120}{60–0.5t}$

To minimize time, we first need to express this as a function of $t$. Cross-multiply:
$t(60–0.5t)=120$
$60t–0.5t^2=120$
$0.5t^2–60t+120=0$

Step 3: Final Answer:
Using the quadratic formula, we find the optimal time.
The solution would yield two time values; choose the one that makes sense in the context of travel time.

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Example 16: Cost Minimization in Manufacturing

Problem:
The cost $C(x)$ of producing $x$ items is given by $C(x)=x^3–15x^2+90x+300$. Find the level of production that minimizes cost.

Answer:
Step 1: Given Data:
The cost function is $C(x)=x^3–15x^2+90x+300$.

Step 2: Solution:
Find the derivative:
$\frac{dC}{dx}=3x^2–30x+90$

Set the derivative equal to zero:
$3x^2–30x+90=0$
$x^2–10x+30=0$

Using the quadratic formula:
$x=\frac{10\pm \sqrt{(-10{)}^2–4(1)(30)}}{2(1)}$
$=\frac{10\pm \sqrt{100–120}}{2}$
$=\frac{10\pm \sqrt{-20}}{2}$

The solutions indicate no real solution for minimum cost.

Step 3: Final Answer:
In practical terms, review cost equations to check feasible production levels.

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Example 17: Revenue Maximization

Problem:
A company sells its product at a price given by $p(x)=100–2x$, where $x$ is the number of products sold. Determine the number of products to sell to maximize revenue.

Answer:
Step 1: Given Data:
Revenue $R$ is given by:
$R=p(x)\cdot x=(100–2x)x=100x–2x^2$

Step 2: Solution:
Find the derivative of the revenue function:
$\frac{dR}{dx}=100–4x$

Set the derivative equal to zero:
$100–4x=0$
$4x=100$
$x=25$

Step 3: Final Answer:
The company should sell 25 products to maximize revenue.

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Example 18: Area of a Rectangle

Problem:
Find the dimensions of a rectangle with a fixed perimeter of 50 meters that maximizes the area.

Answer:
Step 1: Given Data:
Let the length be $l$ meters and the width be $w$ meters. The perimeter is given by:
$2l+2w=50$
Thus,
$l+w=25$
$w=25–l$

Step 2: Solution:
The area $A$ of the rectangle is given by:
$A=l\times w$
Substituting gives:
$A=l(25–l)$
$A=25l–l^2$

To find the maximum area, take the derivative:
$\frac{dA}{dl}=25–2l$

Set the derivative equal to zero:
$25–2l=0$
$2l=25$
$l=12.5$

Substituting back for $w$:
$w=25–12.5=12.5$

Step 3: Final Answer:
The dimensions of the rectangle that maximize the area are $l=12.5$ meters and $w=12.5$ meters.

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Example 19: Minimizing the Surface Area of a Cylinder

Problem:
A cylinder has a volume of 500 cm³. What radius and height will minimize the surface area?

Answer:
Step 1: Given Data:
Let the radius be $r$ cm and the height be $h$ cm. The volume is given by:
$V=\pi r^{2}h=500$

Step 2: Solution:
From the volume equation, solve for $h$:
$h=\frac{500}{\pi r^2}$

The surface area $A$ of the cylinder is:
$A=2\pi r^2+2\pi rh$
Substituting for $h$:
$A=2\pi r^2+2\pi r\cdot \frac{500}{\pi r^2}$
$A=2\pi r^2+\frac{1000}{r}$

To minimize the surface area, take the derivative of $A$:
$\frac{dA}{dr}=4\pi r–\frac{1000}{r^2}$

Set the derivative equal to zero:
$4\pi r=\frac{1000}{r^2}$
$4\pi r^3=1000$
$r^3=\frac{1000}{4\pi }$
$r\approx 5.42\text{ cm}$

Now substitute to find $h$:
$h=\frac{500}{\pi (5.42{)}^2}$

Step 3: Final Answer:
The radius that minimizes surface area is approximately $r\approx 5.42$ cm.

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Example 20: Profit Maximization in Production

Problem:
A firm’s profit function is given by $P(x)=3x^2–24x+60$. Determine the level of production that maximizes profit.

Answer:
Step 1: Given Data:
The profit function is $P(x)=3x^2–24x+60$.

Step 2: Solution:
First, find the derivative:
$\frac{dP}{dx}=6x–24$

Set the derivative equal to zero:
$6x–24=0$
$6x=24$
$x=4$

To confirm it’s a maximum, check the second derivative:
$\frac{d^{2}P}{d{x}^2}=6$

Since the second derivative is positive, $x=4$ gives a maximum.

Step 3: Final Answer:
The firm should produce 4 units to maximize profit.

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Example 21: Maximum Area of a Triangle

Problem:
Find the dimensions of a triangle with a fixed perimeter of 30 meters that maximizes the area.

Answer:
Step 1: Given Data:
Let the sides of the triangle be $a$, $b$, and $c$. The perimeter is given by:
$a+b+c=30$
For maximization, assume $a=b=c$. Thus, $3a=30$ or $a=10$ meters.

Step 2: Solution:
The area $A$ of an equilateral triangle is given by:
$A=\frac{\sqrt{3}}{4}{a}^2$
Substituting for $a$:
$A=\frac{\sqrt{3}}{4}(10{)}^2=25\sqrt{3}{\text{ m}}^2$

Step 3: Final Answer:
The dimensions that maximize the area are $10$ meters for each side.

---

Example 22: Minimizing the Length of a Fence

Problem:
A farmer wants to fence a rectangular area with a fixed area of 200 square meters. What dimensions will minimize the length of the fence?

Answer:
Step 1: Given Data:
Let $l$ be the length and $w$ be the width. The area is given by:
$A=lw=200$
Thus, $w=\frac{200}{l}$.

Step 2: Solution:
The perimeter $P$ (length of the fence) is:
$P=2l+2w=2l+2\left(\frac{200}{l}\right)$
$P=2l+\frac{400}{l}$

To minimize the perimeter, take the derivative:
$\frac{dP}{dl}=2–\frac{400}{l^2}$

Set the derivative equal to zero:
$2–\frac{400}{l^2}=0$
$2l^2=400$
$l^2=200$
$l=\sqrt{200}\approx 14.14\text{ m}$

Now substitute back to find $w$:
$w=\frac{200}{\sqrt{200}}\approx 14.14\text{ m}$

Step 3: Final Answer:
The dimensions that minimize the length of the fence are approximately $14.14$ meters by $14.14$ meters.

---

Example 23: Maximizing a Cylinder’s Volume

Problem:
Find the radius and height of a cylinder that maximizes the volume when the surface area is fixed at 300 square centimeters.

Answer:
Step 1: Given Data:
The surface area of the cylinder is given by:
$A=2\pi r^2+2\pi rh=300$
The volume is:
$V=\pi r^{2}h$.

Step 2: Solution:
From the surface area equation, solve for $h$:
$h=\frac{300–2\pi r^2}{2\pi r}$

Substitute this into the volume formula:
$V=\pi r^2\cdot \frac{300–2\pi r^2}{2\pi r}$
$=\frac{300r}{2}–\pi r^2$
$=150r–\pi r^3$

Now, take the derivative of $V$:
$\frac{dV}{dr}=150–3\pi r^2$

Set the derivative equal to zero:
$150–3\pi r^2=0$
$3\pi r^2=150$
$r^2=\frac{150}{3\pi }$
$r=\sqrt{\frac{50}{\pi }}$

Substituting back to find $h$:
$h=\frac{300–2\pi r^2}{2\pi r}$
$h=\frac{300–2(50)}{2\pi r}=\frac{200}{2\pi r}$
$h=\frac{100}{\pi r}$

Step 3: Final Answer:
The dimensions can be calculated for $r$ and $h$ using the derived values.

---

Example 24: Minimizing Cost in a Factory

Problem:
A factory’s cost function is $C(x)=4x^2+12x+100$. Determine the production level that minimizes costs.

Answer:
Step 1: Given Data:
The cost function is $C(x)=4x^2+12x+100$.

Step 2: Solution:
First, find the derivative:
$\frac{dC}{dx}=8x+12$

Set the derivative equal to zero:
$8x+12=0$
$8x=-12$
$x=-\frac{12}{8}=-\frac{3}{2}$

Since a negative production level is not feasible, evaluate the cost function at production levels within practical limits.

Step 3: Final Answer:
Analyze cost across a reasonable range of $x$.

---

Example 25: Revenue Optimization

Problem:
A company’s revenue function is $R(x)=50x–2x^2$. Find the number of units sold to maximize revenue.

Answer:
Step 1: Given Data:
The revenue function is $R(x)=50x–2x^2$.

Step 2: Solution:
Find the derivative:
$\frac{dR}{dx}=50–4x$

Set the derivative equal to zero:
$50–4x=0$
$4x=50$
$x=12.5$

To confirm it’s a maximum, take the second derivative:
$\frac{d^{2}R}{d{x}^2}=-4$

Since the second derivative is negative, $x=12.5$ gives a maximum.

Step 3: Final Answer:
The company should sell 12.5 units to maximize revenue.

---

Example 26: Cost Function Optimization

Problem:
A cost function is defined as $C(x)=3x^2–24x+60$. What production level minimizes the cost?

Answer:
Step 1: Given Data:
The cost function is $C(x)=3x^2–24x+60$.

Step 2: Solution:
First, find the derivative:
$\frac{dC}{dx}=6x–24$

Set the derivative equal to zero:
$6x–24=0$
$6x=24$
$x=4$

Check the second derivative:
$\frac{d^{2}C}{d{x}^2}=6$

Since the second derivative is positive, $x=4$ is a minimum.

Step 3: Final Answer:
The production level that minimizes cost is $x=4$ units.

---

Example 27: Finding Maximum Height of a Projectile

Problem:
A projectile is launched with a height given by the equation $h(t)=-16t^2+64t+48$. What is the maximum height reached?

Answer:
Step 1: Given Data:
The height function is $h(t)=-16t^2+64t+48$.

Step 2: Solution:
Find the derivative:
$\frac{dh}{dt}=-32t+64$

Set the derivative equal to zero:
$-32t+64=0$
$32t=64$
$t=2$

Substituting $t=2$ back into the height function:
$h(2)=-16(2{)}^2+64(2)+48$
$=-64+128+48$
$=112$

Step 3: Final Answer:
The maximum height reached is $112$ feet.

---

Example 28: Minimizing Material Usage

Problem:
A cylindrical container with a volume of 2000 cm³ is to be made. What radius minimizes the amount of material used?

Answer:
Step 1: Given Data:
The volume is given by:
$V=\pi r^{2}h=2000$
From this, we can express $h$:
$h=\frac{2000}{\pi r^2}$

Step 2: Solution:
The surface area $A$ is given by:
$A=2\pi r^2+2\pi rh$
Substituting $h$:
$A=2\pi r^2+2\pi r\left(\frac{2000}{\pi r^2}\right)$
$=2\pi r^2+\frac{4000}{r}$

Differentiate $A$:
$\frac{dA}{dr}=4\pi r–\frac{4000}{r^2}$

Set the derivative equal to zero:
$4\pi r=\frac{4000}{r^2}$
$4\pi r^3=4000$
$r^3=\frac{1000}{\pi }$
$r=\sqrt[3]{\frac{1000}{\pi }}$

Step 3: Final Answer:
The radius that minimizes material usage is approximately $r\approx 6.83$ cm.

---

Example 29: Finding Maximum Volume

Problem:
A box is to be made from a rectangular piece of cardboard by cutting out squares from the corners. If the cardboard is 20 cm by 30 cm and the squares cut are of size $x$, find the value of $x$ that maximizes the volume of the box.

Answer:
Step 1: Given Data:
The volume $V$ of the box is given by:
$V=x(20–2x)(30–2x)$

Step 2: Solution:
Expanding the volume formula:
$V=x(600–100x–60x+4x^2)$
$=600x–160x^2+4x^3$

Differentiate:
$\frac{dV}{dx}=600–320x+12x^2$

Set the derivative equal to zero:
$12x^2–320x+600=0$
Using the quadratic formula:
$x=\frac{-b\pm \sqrt{b^2–4ac}}{2a}$
$x=\frac{320\pm \sqrt{(-320{)}^2–4\cdot 12\cdot 600}}{2\cdot 12}$

Solve for $x$.

Step 3: Final Answer:
Calculate $x$ to find the optimal value.

---

Example 30: Area of a Triangle Optimization

Problem:
A triangle has a base of 10 meters. What height maximizes the area of the triangle?

Answer:
Step 1: Given Data:
Area of the triangle is given by:
$A=\frac{1}{2}\cdot base\cdot height$
So,
$A=\frac{1}{2}\cdot 10\cdot h=5h$

Step 2: Solution:
Since area increases with height, maximize $h$ subject to constraints of practical dimensions.

Step 3: Final Answer:
Area is maximized when $h$ is at practical maximum based on material used.

---

Example 31: Profit from Selling Price

Problem:
The profit function is $P(x)=100x–5x^2$ where $x$ is the number of units sold. Find the selling price that maximizes profit.

Answer:
Step 1: Given Data:
$P(x)=100x–5x^2$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=100–10x$

Set the derivative to zero:
$100–10x=0$
$10x=100$
$x=10$

Step 3: Final Answer:
The selling price should be set to maximize when $x=10$ units.

---

Example 32: Cost Minimization in Shipping

Problem:
A shipping company wants to minimize costs while maintaining a certain volume. The cost function is $C(x)=20x+\frac{500}{x}$, where $x$ is the length of a side of the box. Find the dimension that minimizes costs.

Answer:
Step 1: Given Data:
The cost function is $C(x)=20x+\frac{500}{x}$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=20–\frac{500}{x^2}$

Set the derivative to zero:
$20–\frac{500}{x^2}=0$
$20x^2=500$
$x^2=25$
$x=5$

Step 3: Final Answer:
The dimension that minimizes cost is $5$ meters.

---

Example 33: Finding Maximum Revenue from Sales

Problem:
A sales function is defined as $R(x)=60x–4x^2$. Find the level of sales that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=60x–4x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=60–8x$

Set the derivative to zero:
$60–8x=0$
$8x=60$
$x=7.5$

Step 3: Final Answer:
Revenue is maximized at a sales level of $7.5$ units.

---

Example 34: Minimizing Water Usage in a Garden

Problem:
A gardener has a rectangular plot of land with a fixed area of 400 square meters. What dimensions minimize the perimeter, thereby reducing water usage?

Answer:
Step 1: Given Data:
Let the length be $l$ and the width be $w$.
$A=lw=400$ implies $w=\frac{400}{l}$.

Step 2: Solution:
Perimeter:
$P=2l+2w=2l+2\left(\frac{400}{l}\right)$
$=2l+\frac{800}{l}$

Differentiate:
$\frac{dP}{dl}=2–\frac{800}{l^2}$

Set the derivative to zero:
$2–\frac{800}{l^2}=0$
$2l^2=800$
$l^2=400$
$l=20$
Now find $w$:
$w=\frac{400}{20}=20$

Step 3: Final Answer:
The dimensions that minimize the perimeter are $20$ meters by $20$ meters.

---

Example 35: Maximizing Efficiency

Problem:
A company has an efficiency function given by $E(x)=-2x^2+40x+150$. What level of production maximizes efficiency?

Answer:
Step 1: Given Data:
$E(x)=-2x^2+40x+150$.

Step 2: Solution:
Differentiate:
$\frac{dE}{dx}=-4x+40$

Set the derivative to zero:
$-4x+40=0$
$4x=40$
$x=10$

Step 3: Final Answer:
The production level that maximizes efficiency is $x=10$.

---

Example 36: Cost Function with Constraints

Problem:
A manufacturing company’s cost function is $C(x)=3x^2+12x+18$. Determine the production level that minimizes the cost.

Answer:
Step 1: Given Data:
The cost function is $C(x)=3x^2+12x+18$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=6x+12$

Set the derivative to zero:
$6x+12=0$
$6x=-12$
$x=-2$

Since a negative production level is not feasible, check costs over practical values of $x$.

Step 3: Final Answer:
Optimize around $x=0$ to determine the least cost.

---

Example 37: Energy Consumption Minimization

Problem:
An energy consumption function is defined as $E(x)=x^3–15x^2+72x$. What consumption level minimizes energy usage?

Answer:
Step 1: Given Data:
$E(x)=x^3–15x^2+72x$.

Step 2: Solution:
Differentiate:
$\frac{dE}{dx}=3x^2–30x+72$

Set the derivative to zero and solve for $x$:
$3x^2–30x+72=0$
Using the quadratic formula:
$x=\frac{-b\pm \sqrt{b^2–4ac}}{2a}$

Step 3: Final Answer:
Calculate $x$ for the minimum energy consumption.

---

Example 38: Balancing a Profit Function

Problem:
A profit function is represented as $P(x)=-x^2+40x–300$. Determine the quantity of production that maximizes profit.

Answer:
Step 1: Given Data:
$P(x)=-x^2+40x–300$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-2x+40$

Set the derivative to zero:
$-2x+40=0$
$2x=40$
$x=20$

Step 3: Final Answer:
Maximum profit occurs at $x=20$ units.

---

Example 39: Finding Minimum Cost in Manufacturing

Problem:
A company has a cost function given by $C(x)=5x^2+30x+250$. Find the production level that minimizes cost.

Answer:
Step 1: Given Data:
$C(x)=5x^2+30x+250$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=10x+30$

Set the derivative to zero:
$10x+30=0$
$10x=-30$
$x=-3$

Evaluate costs at feasible levels of $x$.

Step 3: Final Answer:
Analyze cost around practical production levels.

---

Example 40: Supply Function Maximization

Problem:
A company’s supply function is $S(x)=3x^2–12x+18$. Find the production level that maximizes supply.

Answer:
Step 1: Given Data:
$S(x)=3x^2–12x+18$.

Step 2: Solution:
Differentiate:
$\frac{dS}{dx}=6x–12$

Set the derivative to zero:
$6x–12=0$
$6x=12$
$x=2$

Step 3: Final Answer:
Maximum supply occurs at $x=2$ units.

---

Example 41: Minimizing Advertising Cost

Problem:
An advertising cost function is defined as $C(x)=200+3x+0.01x^2$, where $x$ is the number of advertisements. Find the optimal number of advertisements that minimizes cost.

Answer:
Step 1: Given Data:
$C(x)=200+3x+0.01x^2$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=3+0.02x$

Set the derivative to zero:
$3+0.02x=0$
$0.02x=-3$
$x=-150$

Check costs at feasible levels of advertisements.

Step 3: Final Answer:
Determine practical $x$ to minimize advertising cost.

---

Example 42: Profit Function in Retail

Problem:
A retail store’s profit function is $P(x)=-4x^2+80x–200$. Determine the number of items that maximizes profit.

Answer:
Step 1: Given Data:
$P(x)=-4x^2+80x–200$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-8x+80$

Set the derivative to zero:
$-8x+80=0$
$8x=80$
$x=10$

Step 3: Final Answer:
Maximized profit occurs at $x=10$ items.

---

Example 43: Optimization of Travel Costs

Problem:
A travel cost function is given by $C(x)=50x+0.2x^2$, where $x$ is the number of miles. Find the optimal travel distance that minimizes cost.

Answer:
Step 1: Given Data:
$C(x)=50x+0.2x^2$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=50+0.4x$

Set the derivative to zero:
$50+0.4x=0$
$0.4x=-50$
$x=-125$

Evaluate costs around practical travel distances.

Step 3: Final Answer:
Assess distance choices for cost minimization.

---

Example 44: Minimizing Distance Traveled

Problem:
A delivery service wants to minimize the distance traveled, modeled by the function $D(x)=x^2–4x+10$. Determine the optimal distance.

Answer:
Step 1: Given Data:
$D(x)=x^2–4x+10$.

Step 2: Solution:
Differentiate:
$\frac{dD}{dx}=2x–4$

Set the derivative to zero:
$2x–4=0$
$2x=4$
$x=2$

Step 3: Final Answer:
Minimum distance occurs at $x=2$ miles.

---

Example 45: Cost Function Analysis

Problem:
A cost function is defined as $C(x)=10x^2+40x+50$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=10x^2+40x+50$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=20x+40$

Set the derivative to zero:
$20x+40=0$
$20x=-40$
$x=-2$

Evaluate for practical production levels.

Step 3: Final Answer:
Assess practical production levels to minimize cost.

---

Example 46: Maximizing a Product’s Profit

Problem:
A product’s profit function is $P(x)=-3x^2+150x–900$. What is the maximum profit?

Answer:
Step 1: Given Data:
$P(x)=-3x^2+150x–900$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-6x+150$

Set the derivative to zero:
$-6x+150=0$
$6x=150$
$x=25$

Step 3: Final Answer:
Maximum profit occurs at $x=25$ products sold.

---

Example 47: Minimizing Fuel Consumption

Problem:
A car’s fuel consumption can be modeled by the function $C(x)=5x^2+3x+15$. Determine the speed that minimizes fuel consumption.

Answer:
Step 1: Given Data:
$C(x)=5x^2+3x+15$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=10x+3$

Set the derivative to zero:
$10x+3=0$
$10x=-3$
$x=-0.3$

Evaluate for practical speeds for fuel efficiency.

Step 3: Final Answer:
Examine practical values for fuel optimization.

---

Example 48: Optimizing Material Costs

Problem:
A company’s cost function for material is $C(x)=2x^2+100$. Find the optimal material amount that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=2x^2+100$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=4x$

Set the derivative to zero:
$4x=0$
$x=0$

Step 3: Final Answer:
Cost is minimized at $x=0$.

---

Example 49: Minimizing Waste in Production

Problem:
A company produces waste based on the function $W(x)=4x^2+20x+40$. Find the production level that minimizes waste.

Answer:
Step 1: Given Data:
$W(x)=4x^2+20x+40$.

Step 2: Solution:
Differentiate:
$\frac{dW}{dx}=8x+20$

Set the derivative to zero:
$8x+20=0$
$8x=-20$
$x=-\frac{20}{8}=-2.5$

Evaluate for practical production levels.

Step 3: Final Answer:
Identify practical levels to minimize waste.

---

Example 50: Revenue Maximization

Problem:
A company’s revenue function is $R(x)=100x–4x^2$. Find the number of items that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=100x–4x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=100–8x$

Set the derivative to zero:
$100–8x=0$
$8x=100$
$x=12.5$

Step 3: Final Answer:
Revenue is maximized at $x=12.5$ items sold.

---

Example 51: Cost Analysis in Construction

Problem:
A construction company’s cost function is $C(x)=10x^2–40x+300$. Find the minimum cost.

Answer:
Step 1: Given Data:
$C(x)=10x^2–40x+300$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=20x–40$

Set the derivative to zero:
$20x–40=0$
$20x=40$
$x=2$

Step 3: Final Answer:
Minimum cost occurs at $x=2$ units.

---

Example 52: Area Optimization

Problem:
A rectangular garden has a fixed perimeter of 60 meters. Find the dimensions that maximize the area.

Answer:
Step 1: Given Data:
Let $l$ be the length and $w$ the width. The perimeter is:
$2l+2w=60$
Thus, $l+w=30$.

Step 2: Solution:
The area is given by:
$A=lw=l(30–l)=30l–l^2$

Differentiate the area function:
$\frac{dA}{dl}=30–2l$

Set the derivative equal to zero:
$30–2l=0$
$2l=30$
$l=15$
Then, $w=30–l=15$.

Step 3: Final Answer:
The dimensions that maximize the area are $15$ meters by $15$ meters.

---

Example 53: Revenue Optimization

Problem:
The revenue function is $R(x)=150x–5x^2$. What number of units should be sold to maximize revenue?

Answer:
Step 1: Given Data:
$R(x)=150x–5x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=150–10x$

Set the derivative equal to zero:
$150–10x=0$
$10x=150$
$x=15$

Step 3: Final Answer:
To maximize revenue, sell $15$ units.

---

Example 54: Finding Optimal Pricing Strategy

Problem:
A pricing function is given by $P(x)=-3x^2+60x$. Determine the price point that maximizes revenue.

Answer:
Step 1: Given Data:
$P(x)=-3x^2+60x$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-6x+60$

Set the derivative equal to zero:
$-6x+60=0$
$6x=60$
$x=10$

Step 3: Final Answer:
The optimal pricing strategy occurs at $x=10$.

---

Example 55: Minimizing Waste in Packaging

Problem:
The waste function in a factory is represented by $W(x)=x^3–15x^2+75x$. Find the production level that minimizes waste.

Answer:
Step 1: Given Data:
$W(x)=x^3–15x^2+75x$.

Step 2: Solution:
Differentiate:
$\frac{dW}{dx}=3x^2–30x+75$

Set the derivative to zero:
$3x^2–30x+75=0$
Using the quadratic formula:
$x=\frac{-b\pm \sqrt{b^2–4ac}}{2a}$

Step 3: Final Answer:
Calculate $x$ values for minimal waste.

---

Example 56: Profit Maximization for a Business

Problem:
A business’s profit function is $P(x)=-2x^2+20x–40$. Determine the production level that maximizes profit.

Answer:
Step 1: Given Data:
$P(x)=-2x^2+20x–40$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-4x+20$

Set the derivative to zero:
$-4x+20=0$
$4x=20$
$x=5$

Step 3: Final Answer:
Maximum profit occurs at $x=5$ units.

---

Example 57: Cost Function of a Business

Problem:
A cost function is defined by $C(x)=3x^2+24x+18$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=3x^2+24x+18$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=6x+24$

Set the derivative to zero:
$6x+24=0$
$6x=-24$
$x=-4$

Evaluate costs at practical levels of $x$.

Step 3: Final Answer:
Check for minimized cost across feasible production levels.

---

Example 58: Volume Optimization of a Box

Problem:
A box’s volume function is given by $V(x)=x^2(10–x)$. Find the dimension that maximizes the volume.

Answer:
Step 1: Given Data:
$V(x)=x^2(10–x)$.

Step 2: Solution:
Differentiate:
$\frac{dV}{dx}=2x(10–x)–x^2$
$=20x–3x^2$

Set the derivative equal to zero:
$20x–3x^2=0$
$x(20–3x)=0$
Thus, $x=0$ or $x=\frac{20}{3}\approx 6.67$.

Step 3: Final Answer:
Volume is maximized at dimensions around $x\approx 6.67$.

---

Example 59: Optimal Strategy for Selling Price

Problem:
A profit function is $P(x)=-2x^2+50x–200$. What selling price maximizes profit?

Answer:
Step 1: Given Data:
$P(x)=-2x^2+50x–200$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-4x+50$

Set the derivative to zero:
$-4x+50=0$
$4x=50$
$x=12.5$

Step 3: Final Answer:
Maximized profit occurs at a selling price of $x=12.5$ units.

---

Example 60: Cost Minimization in Production

Problem:
The cost function is $C(x)=5x^2–40x+100$. Find the production level that minimizes cost.

Answer:
Step 1: Given Data:
$C(x)=5x^2–40x+100$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=10x–40$

Set the derivative to zero:
$10x–40=0$
$10x=40$
$x=4$

Step 3: Final Answer:
Cost is minimized at $x=4$ units.

---

Example 61: Area Optimization for a Rectangle

Problem:
A rectangular area has a fixed perimeter of 50 meters. What dimensions maximize the area?

Answer:
Step 1: Given Data:
Let the length be $l$ and width be $w$.
$2l+2w=50$
So, $l+w=25$.

Step 2: Solution:
Area is given by:
$A=lw=l(25–l)=25l–l^2$

Differentiate:
$\frac{dA}{dl}=25–2l$

Set the derivative equal to zero:
$25–2l=0$
$2l=25$
$l=12.5$
Then, $w=25–12.5=12.5$.

Step 3: Final Answer:
Dimensions that maximize area are $12.5$ meters by $12.5$ meters.

---

Example 62: Finding Optimal Production Levels

Problem:
A factory’s production function is given by $P(x)=-x^2+100x$. Determine the production level that maximizes output.

Answer:
Step 1: Given Data:
$P(x)=-x^2+100x$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-2x+100$

Set the derivative equal to zero:
$-2x+100=0$
$2x=100$
$x=50$

Step 3: Final Answer:
Maximum output occurs at $x=50$ units.

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Example 63: Optimizing Distribution of Resources

Problem:
A company wants to minimize the cost of production given by $C(x)=8x^2–48x+64$. What production level minimizes cost?

Answer:
Step 1: Given Data:
$C(x)=8x^2–48x+64$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=16x–48$

Set the derivative equal to zero:
$16x–48=0$
$16x=48$
$x=3$

Step 3: Final Answer:
Cost is minimized at $x=3$ units.

---

Example 64: Revenue Maximization from Sales

Problem:
A sales revenue function is $R(x)=200x–5x^2$. Find the number of items to maximize revenue.

Answer:
Step 1: Given Data:
$R(x)=200x–5x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=200–10x$

Set the derivative equal to zero:
$200–10x=0$
$10x=200$
$x=20$

Step 3: Final Answer:
Maximized revenue occurs when selling $20$ items.

---

Example 65: Cost of Manufacturing Process

Problem:
A manufacturing cost function is defined by $C(x)=4x^2+50$. What production level minimizes costs?

Answer:
Step 1: Given Data:
$C(x)=4x^2+50$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=8x$

Set the derivative equal to zero:
$8x=0$
$x=0$

Step 3: Final Answer:
Minimum cost occurs at $x=0$ units.

---

Example 66: Finding Maximum Production Levels

Problem:
A production function is given by $P(x)=10x–2x^2$. Determine the number of units that maximizes production.

Answer:
Step 1: Given Data:
$P(x)=10x–2x^2$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=10–4x$

Set the derivative equal to zero:
$10–4x=0$
$4x=10$
$x=2.5$

Step 3: Final Answer:
Maximum production occurs at $x=2.5$ units.

---

Example 67: Minimizing Costs in a Factory

Problem:
A factory’s cost function is defined as $C(x)=2x^2+30x+100$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=2x^2+30x+100$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=4x+30$

Set the derivative equal to zero:
$4x+30=0$
$4x=-30$
$x=-\frac{15}{2}$

Evaluate costs at feasible levels of production.

Step 3: Final Answer:
Assess production levels for minimized costs.

---

Example 68: Area Maximization for a Rectangular Plot

Problem:
A rectangular plot has a fixed perimeter of 40 meters. What dimensions maximize the area?

Answer:
Step 1: Given Data:
Let the length be $l$ and width be $w$.
$2l+2w=40$
Thus, $l+w=20$.

Step 2: Solution:
Area is given by:
$A=lw=l(20–l)=20l–l^2$

Differentiate:
$\frac{dA}{dl}=20–2l$

Set the derivative equal to zero:
$20–2l=0$
$2l=20$
$l=10$
Then, $w=20–10=10$.

Step 3: Final Answer:
The dimensions that maximize area are $10$ meters by $10$ meters.

---

Example 69: Finding Minimum Cost for Product Production

Problem:
A product’s cost function is $C(x)=3x^2+6x+50$. Determine the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=3x^2+6x+50$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=6x+6$

Set the derivative equal to zero:
$6x+6=0$
$6x=-6$
$x=-1$

Evaluate costs at feasible levels of production.

Step 3: Final Answer:
Analyze production levels to minimize costs.

---

Example 70: Maximizing Advertising Effectiveness

Problem:
An advertising effectiveness function is $E(x)=-5x^2+70x$. Find the level of advertising that maximizes effectiveness.

Answer:
Step 1: Given Data:
$E(x)=-5x^2+70x$.

Step 2: Solution:
Differentiate:
$\frac{dE}{dx}=-10x+70$

Set the derivative equal to zero:
$-10x+70=0$
$10x=70$
$x=7$

Step 3: Final Answer:
Maximum effectiveness occurs at $x=7$ advertisements.

---

Example 71: Finding Maximum Volume of a Box

Problem:
A box is constructed from a square piece of cardboard by cutting out squares from the corners. If the cardboard is 12 cm by 12 cm, find the side length of the squares cut to maximize the volume of the box.

Answer:
Step 1: Given Data:
Let $x$ be the side length of the squares cut. The volume $V$ of the box is given by:
$V=x(12–2x)(12–2x)=x(144–48x+4x^2)$
$=4x^3–48x^2+144x$

Step 2: Solution:
Differentiate:
$\frac{dV}{dx}=12x^2–96x+144$

Set the derivative equal to zero:
$12x^2–96x+144=0$
Dividing by 12:
$x^2–8x+12=0$

Using the quadratic formula:
$x=\frac{8\pm \sqrt{(-8{)}^2–4\cdot 1\cdot 12}}{2\cdot 1}$

Step 3: Final Answer:
Calculate $x$ values for maximum volume.

---

Example 72: Maximizing Return on Investment

Problem:
A return function is defined by $R(x)=150x–3x^2$. Find the investment level that maximizes return.

Answer:
Step 1: Given Data:
$R(x)=150x–3x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=150–6x$

Set the derivative equal to zero:
$150–6x=0$
$6x=150$
$x=25$

Step 3: Final Answer:
Maximum return occurs at an investment level of $x=25$.

---

Example 73: Optimizing Factory Production

Problem:
A factory’s production function is given by $P(x)=-x^2+20x$. Find the production level that maximizes output.

Answer:
Step 1: Given Data:
$P(x)=-x^2+20x$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-2x+20$

Set the derivative equal to zero:
$-2x+20=0$
$2x=20$
$x=10$

Step 3: Final Answer:
Maximum production occurs at $x=10$ units.

---

Example 74: Minimizing Waste in Production

Problem:
A waste function is given by $W(x)=x^2–4x+4$. Find the production level that minimizes waste.

Answer:
Step 1: Given Data:
$W(x)=x^2–4x+4$.

Step 2: Solution:
Differentiate:
$\frac{dW}{dx}=2x–4$

Set the derivative equal to zero:
$2x–4=0$
$2x=4$
$x=2$

Step 3: Final Answer:
Minimum waste occurs at $x=2$ units.

---

Example 75: Finding Maximum Production for a Product

Problem:
The production function is defined by $P(x)=-3x^2+30x–72$. Find the production level that maximizes output.

Answer:
Step 1: Given Data:
$P(x)=-3x^2+30x–72$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-6x+30$

Set the derivative equal to zero:
$-6x+30=0$
$6x=30$
$x=5$

Step 3: Final Answer:
Maximum production occurs at $x=5$ units.

---

Example 76: Cost Minimization in Shipping

Problem:
A shipping cost function is given by $C(x)=2x^2+8x+50$. Find the shipping level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=2x^2+8x+50$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=4x+8$

Set the derivative equal to zero:
$4x+8=0$
$4x=-8$
$x=-2$

Evaluate for practical shipping levels.

Step 3: Final Answer:
Assess practical levels for minimized shipping costs.

---

Example 77: Revenue Optimization for a Service

Problem:
A service’s revenue function is represented as $R(x)=100x–3x^2$. Determine the service level that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=100x–3x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=100–6x$

Set the derivative equal to zero:
$100–6x=0$
$6x=100$
$x=\frac{100}{6}\approx 16.67$

Step 3: Final Answer:
Revenue is maximized at a service level of approximately $16.67$.

---

Example 78: Profit Maximization in a Business Model

Problem:
A business’s profit function is $P(x)=-2x^2+40x–150$. Find the quantity that maximizes profit.

Answer:
Step 1: Given Data:
$P(x)=-2x^2+40x–150$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-4x+40$

Set the derivative equal to zero:
$-4x+40=0$
$4x=40$
$x=10$

Step 3: Final Answer:
Maximum profit occurs at $x=10$ units.

---

Example 79: Cost Function Analysis for a Retail Store

Problem:
A retail store’s cost function is given by $C(x)=5x^2–10x+100$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=5x^2–10x+100$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=10x–10$

Set the derivative equal to zero:
$10x–10=0$
$10x=10$
$x=1$

Step 3: Final Answer:
Minimum cost occurs at $x=1$ unit.

---

Example 80: Area Optimization for a Circular Plot

Problem:
A circular garden has a fixed area of 100 square meters. What radius maximizes the area?

Answer:
Step 1: Given Data:
Let $A=\pi r^2=100$.

Step 2: Solution:
Solve for $r$:
$r^2=\frac{100}{\pi }$
$r=\sqrt{\frac{100}{\pi }}$

Step 3: Final Answer:
The radius that maximizes area is approximately $r\approx 5.64$ meters.

---

Example 81: Finding Optimal Production Levels

Problem:
A company’s production function is given by $P(x)=-2x^2+30x$. Find the production level that maximizes output.

Answer:
Step 1: Given Data:
$P(x)=-2x^2+30x$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-4x+30$

Set the derivative equal to zero:
$-4x+30=0$
$4x=30$
$x=7.5$

Step 3: Final Answer:
Maximum output occurs at $x=7.5$ units.

---

Example 82: Minimizing Costs in a Factory

Problem:
A factory’s cost function is $C(x)=6x^2+10x+20$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=6x^2+10x+20$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=12x+10$

Set the derivative to zero:
$12x+10=0$
$12x=-10$
$x=-\frac{5}{6}$

Evaluate for practical levels of production.

Step 3: Final Answer:
Assess practical levels for minimized costs.

---

Example 83: Revenue Maximization for a Business

Problem:
A company’s revenue function is $R(x)=80x–2x^2$. Determine the production level that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=80x–2x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=80–4x$

Set the derivative equal to zero:
$80–4x=0$
$4x=80$
$x=20$

Step 3: Final Answer:
Revenue is maximized when $x=20$ units.

---

Example 84: Minimizing Shipping Costs

Problem:
A shipping cost function is given by $C(x)=3x^2+15$. Find the shipping level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=3x^2+15$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=6x$

Set the derivative to zero:
$6x=0$
$x=0$

Step 3: Final Answer:
Minimum cost occurs at $x=0$.

---

Example 85: Maximizing Profit in a Manufacturing Plant

Problem:
A profit function is defined as $P(x)=-3x^2+60x–200$. What production level maximizes profit?

Answer:
Step 1: Given Data:
$P(x)=-3x^2+60x–200$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-6x+60$

Set the derivative to zero:
$-6x+60=0$
$6x=60$
$x=10$

Step 3: Final Answer:
Maximum profit occurs at $x=10$ units.

---

Example 86: Cost Function Optimization in Production

Problem:
The cost function is defined by $C(x)=7x^2–42x+100$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=7x^2–42x+100$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=14x–42$

Set the derivative to zero:
$14x–42=0$
$14x=42$
$x=3$

Step 3: Final Answer:
Minimum cost occurs at $x=3$ units.

---

Example 87: Optimizing Advertising Strategy

Problem:
An advertising function is given by $A(x)=-2x^2+40x$. Find the level of advertising that maximizes effectiveness.

Answer:
Step 1: Given Data:
$A(x)=-2x^2+40x$.

Step 2: Solution:
Differentiate:
$\frac{dA}{dx}=-4x+40$

Set the derivative to zero:
$-4x+40=0$
$4x=40$
$x=10$

Step 3: Final Answer:
Maximum effectiveness occurs at $x=10$ advertisements.

---

Example 88: Cost Minimization in a Production Plant

Problem:
A cost function is defined by $C(x)=2x^2+8x+50$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=2x^2+8x+50$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=4x+8$

Set the derivative to zero:
$4x+8=0$
$4x=-8$
$x=-2$

Evaluate costs at feasible levels of production.

Step 3: Final Answer:
Assess production levels to minimize costs.

---

Example 89: Revenue Optimization for a New Product

Problem:
A new product’s revenue function is given by $R(x)=150x–3x^2$. Find the production level that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=150x–3x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=150–6x$

Set the derivative to zero:
$150–6x=0$
$6x=150$
$x=25$

Step 3: Final Answer:
Maximized revenue occurs at a production level of $x=25$ units.

---

Example 90: Finding Optimal Pricing for Maximum Revenue

Problem:
The revenue function is defined as $R(x)=200x–4x^2$. Determine the price level that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=200x–4x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=200–8x$

Set the derivative to zero:
$200–8x=0$
$8x=200$
$x=25$

Step 3: Final Answer:
Maximum revenue occurs at a price level of $x=25$.

---

Example 91: Cost Minimization in a Factory

Problem:
A factory’s cost function is $C(x)=4x^2+12x+36$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=4x^2+12x+36$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=8x+12$

Set the derivative to zero:
$8x+12=0$
$8x=-12$
$x=-\frac{3}{2}$

Evaluate for practical production levels.

Step 3: Final Answer:
Analyze production levels to minimize costs.

---

Example 92: Maximizing a Function’s Output

Problem:
A function $f(x)=-3x^2+60x–300$ represents output. Find the production level that maximizes output.

Answer:
Step 1: Given Data:
$f(x)=-3x^2+60x–300$.

Step 2: Solution:
Differentiate:
$\frac{df}{dx}=-6x+60$

Set the derivative to zero:
$-6x+60=0$
$6x=60$
$x=10$

Step 3: Final Answer:
Maximum output occurs at $x=10$.

---

Example 93: Finding Minimum Production Levels

Problem:
A production cost function is given by $C(x)=5x^2+10x+25$. What production level minimizes costs?

Answer:
Step 1: Given Data:
$C(x)=5x^2+10x+25$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=10x+10$

Set the derivative to zero:
$10x+10=0$
$10x=-10$
$x=-1$

Evaluate for feasible production levels.

Step 3: Final Answer:
Assess production levels for minimized costs.

---

Example 94: Revenue Function Maximization

Problem:
A revenue function is defined as $R(x)=100x–2x^2$. Determine the production level that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=100x–2x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=100–4x$

Set the derivative to zero:
$100–4x=0$
$4x=100$
$x=25$

Step 3: Final Answer:
Maximum revenue occurs at a production level of $x=25$.

---

Example 95: Cost Function Analysis in Manufacturing

Problem:
A manufacturing cost function is $C(x)=6x^2–24x+50$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=6x^2–24x+50$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=12x–24$

Set the derivative to zero:
$12x–24=0$
$12x=24$
$x=2$

Step 3: Final Answer:
Minimum cost occurs at $x=2$ units.

---

Example 96: Profit Maximization in a Retail Store

Problem:
A retail store’s profit function is $P(x)=-2x^2+80x–200$. Determine the production level that maximizes profit.

Answer:
Step 1: Given Data:
$P(x)=-2x^2+80x–200$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-4x+80$

Set the derivative to zero:
$-4x+80=0$
$4x=80$
$x=20$

Step 3: Final Answer:
Maximum profit occurs at $x=20$ units.

---

Example 97: Cost Function Minimization in Production

Problem:
A production cost function is $C(x)=2x^2+12x+10$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=2x^2+12x+10$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=4x+12$

Set the derivative to zero:
$4x+12=0$
$4x=-12$
$x=-3$

Evaluate for practical production levels.

Step 3: Final Answer:
Analyze production levels for minimized costs.

---

Example 98: Revenue Maximization for a Business Strategy

Problem:
A business strategy’s revenue function is $R(x)=200x–5x^2$. Find the optimal strategy that maximizes revenue.

Answer:
Step 1: Given Data:
$R(x)=200x–5x^2$.

Step 2: Solution:
Differentiate:
$\frac{dR}{dx}=200–10x$

Set the derivative to zero:
$200–10x=0$
$10x=200$
$x=20$

Step 3: Final Answer:
Maximum revenue occurs at a production strategy of $x=20$.

---

Example 99: Profit Maximization in Retail

Problem:
A retail profit function is defined as $P(x)=-x^2+40x–100$. Determine the production level that maximizes profit.

Answer:
Step 1: Given Data:
$P(x)=-x^2+40x–100$.

Step 2: Solution:
Differentiate:
$\frac{dP}{dx}=-2x+40$

Set the derivative to zero:
$-2x+40=0$
$2x=40$
$x=20$

Step 3: Final Answer:
Maximum profit occurs at $x=20$ units.

---

Example 100: Minimizing Cost in a Production Facility

Problem:
A production cost function is defined as $C(x)=x^2+20x+100$. Find the production level that minimizes costs.

Answer:
Step 1: Given Data:
$C(x)=x^2+20x+100$.

Step 2: Solution:
Differentiate:
$\frac{dC}{dx}=2x+20$

Set the derivative to zero:
$2x+20=0$
$2x=-20$
$x=-10$

Evaluate costs at practical production levels.

Step 3: Final Answer:
Assess production levels for minimized costs.