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## How to Solve Optimization Problems in Calculus

- July 7, 2016
- by Bruce Birkett
- Tags: Calculus , can , optimization , problem solving strategy
- 16 Comments

## Stage I. Develop the function

## Stage II: Maximize or minimize your function

For instance, a few weeks ago you could have gotten this as a standard max/min homework problem:

The minimum surface area occurs when $r = \sqrt[3]{\dfrac{V}{2\pi}}\,. \quad \triangleleft$

## Summary: Problem Solving Strategy

## Stage I: Develop the function.

- Draw a picture of the physical situation. Also note any physical restrictions determined by the physical situation.
- Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
- If necessary, use other given information to rewrite your equation in terms of a single variable.

## Stage II: Maximize or minimize the function.

- Take the derivative of your equation with respect to your single variable. Then find the critical points.
- Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
- Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.
- Finally, check to make sure you have answered the question as asked : Re-read the problem and verify that you are providing the value(s) requested: an x or y value; or coordinates; or a maximum area; or a shortest time; whatever was asked.

- What tips do you have to share about how to solve Optimization problems?
- What questions do you have? Optimization problems can be tricky to start, and we’re happy to help !
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what problems can help to solve optimization

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## Optimization In Calculus How-To w/ 7 Step-by-Step Examples!

// Last Updated: July 25, 2021 - Watch Video //

Has there ever been a time when you wish the day would never end?

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

Or, on the flip side, have you ever felt like the day couldn’t end fast enough?

What do these sentiments have in common?

Both are trying to optimize the situation!

Optimization is the process of finding maximum and minimum values given constraints using calculus .

For example, you’ll be given a situation where you’re asked to find:

It is our job to translate the problem or picture into usable functions to find the extreme values .

## Solving Optimization Problems (Step-by-Step)

Step 2: Substitute our secondary equation into our primary equation and simplify.

Let’s look at a few problems to see how our optimization problem-solving strategies in work.

First, we need to find our primary and secondary equations by translating our problem.

Secondary: \(x \cdot y=192\) Primary: \(x+y=\min\)

See, that wasn’t too bad now, was it?

Okay, let’s now look at a more realistic question.

First, let’s draw a picture and find our primary and secondary equations.

We know that the area of the enclosure is 600, so that would mean:

\begin{equation} A=x \cdot y=600 \end{equation}

Cost Materials Rectangular Enclosure

\begin{equation} C=7(\text { wood })+14(\text { brick })=7(2 y+x)+14 x=14 y+21 x \end{equation}

Alright, so now we’ve acquired our two equations.

Primary: \(C=14 y+21 x\) Secondary: \(x y=600\)

The second derivative is positive at y = 30, so we know that we have a local minimum!

\begin{equation} \begin{array}{l} x y=600, y=30 \\ x(30)=600 \\ x=20 \end{array} \end{equation}

This means that the dimensions of the least costly enclosure are 20 feet long and 30 feet wide.

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## 4.7: Optimization Problems

## Learning Objectives

## Solving Optimization Problems over a Closed, Bounded Interval

## Example \(\PageIndex{1}\): Maximizing the Area of a Garden

Solving this equation for \(y\), we have \(y=100−2x.\) Thus, we can write the area as

\(A(x)=x⋅(100−2x)=100x−2x^2.\)

Therefore, we consider the following problem:

Maximize \(A(x)=100x−2x^2\) over the interval \([0,50].\)

## Exercise \(\PageIndex{1}\)

We need to maximize the function \(A(x)=200x−2x^2\) over the interval \([0,100].\)

The maximum area is \(5000\, \text{ft}^2\).

## Problem-Solving Strategy: Solving Optimization Problems

- Introduce all variables. If applicable, draw a figure and label all variables.
- Determine which quantity is to be maximized or minimized, and for what range of values of the other variables (if this can be determined at this time).
- Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more than one variable.
- Write any equations relating the independent variables in the formula from step \(3\). Use these equations to write the quantity to be maximized or minimized as a function of one variable.
- Identify the domain of consideration for the function in step \(4\) based on the physical problem to be solved.
- Locate the maximum or minimum value of the function from step \(4.\) This step typically involves looking for critical points and evaluating a function at endpoints.

## Example \(\PageIndex{2}\): Maximizing the Volume of a Box

Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize \(V\).

Step 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is

\[V=L⋅W⋅H \nonumber, \nonumber \]

where \(L,\,W,\)and \(H\) are the length, width, and height, respectively.

\[ \begin{align*} V(x) &=(36−2x)(24−2x)x \\[4pt] &=4x^3−120x^2+864x \end{align*}. \nonumber \]

To find the critical points, we need to solve the equation

Dividing both sides of this equation by \(12\), the problem simplifies to solving the equation

Using the quadratic formula, we find that the critical points are

\[V(10−2\sqrt{7})=640+448\sqrt{7}≈1825\,\text{in}^3. \nonumber \]

as shown in the following graph.

## Exercise \(\PageIndex{2}\)

The volume of the box is \(L⋅W⋅H.\)

\(V(x)=x(20−2x)(30−2x).\) The domain is \([0,10]\).

## Example \(\PageIndex{3}\): Minimizing Travel Time

Step 2: The problem is to minimize \(T\).

\(T_{running}=\dfrac{D_{running}}{R_{running}}=\dfrac{x}{8}\),

and the time spent swimming is

\(T_{swimming}=\dfrac{D_{swimming}}{R_{swimming}}=\dfrac{y}{3}\).

Therefore, the total time spent traveling is

\(T=\dfrac{x}{8}+\dfrac{y}{3}\).

\(T(x)=\dfrac{x}{8}+\dfrac{\sqrt{(6−x)^2+4}}{3}\).

\[\dfrac{1}{8}=\dfrac{6−x}{3\sqrt{(6−x)^2+4}} \label{ex3eq1} \]

\[3\sqrt{(6−x)^2+4}=8(6−x). \label{ex3eq2} \]

\[9[(6−x)^2+4]=64(6−x)^2,\nonumber \]

We conclude that if \(x\) is a critical point, then \(x\) satisfies

\[(x−6)^2=\dfrac{36}{55}. \nonumber \]

[Note that since we are squaring, \( (x-6)^2 = (6-x)^2.\)]

Therefore, the possibilities for critical points are

\[x=6±\dfrac{6}{\sqrt{55}}.\nonumber \]

\[T(6−6/\sqrt{55})≈1.368\,\text{h}. \nonumber \]

Therefore, we conclude that \(T\) has a local minimum at \(x≈5.19\) mi.

## Exercise \(\PageIndex{3}\)

The time \(T=T_{running}+T_{swimming}.\)

\(T(x)=\dfrac{x}{6}+\dfrac{\sqrt{(15−x)^2+1}}{2.5} \)

## Example \(\PageIndex{4}\): Maximizing Revenue

Step 2: The problem is to maximize \(R.\)

\[ \begin{align*} R(p) &=n×p \\[4pt] &=(1000−5p)p \\[4pt] &=−5p^2+1000p.\end{align*}\]

## Exercise \(\PageIndex{4}\)

\(R(p)=n×p,\) where \(n\) is the number of cars rented and \(p\) is the price charged per car.

The company should charge \($75\) per car per day.

## Example \(\PageIndex{5}\): Maximizing the Area of an Inscribed Rectangle

A rectangle is to be inscribed in the ellipse

\[\dfrac{x^2}{4}+y^2=1. \nonumber \]

What should the dimensions of the rectangle be to maximize its area? What is the maximum area?

Step 2: The problem is to maximize \(A\).

Step 3: The area of the rectangle is \(A=LW.\)

\(A=LW=(2x)(2y)=4x\sqrt{1-\dfrac{x^2}{4}}=2x\sqrt{4−x^2}\)

Therefore, the maximum must occur at a critical point. Taking the derivative of \(A(x)\), we obtain

\[\dfrac{8−4x^2}{\sqrt{4−x^2}}=0, \label{ex5eq1} \]

\[y=\sqrt{1−\dfrac{(\sqrt{2})^2}{4}}=\sqrt{1−\dfrac{1}{2}}=\dfrac{1}{\sqrt{2}}.\nonumber \]

## Exercise \(\PageIndex{5}\)

\(A(x)=4x\sqrt{1−x^2}.\) The domain of consideration is \([0,1]\).

## Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded

## Example \(\PageIndex{6}\): Minimizing Surface Area

Step 2: We need to minimize the surface area. Therefore, we need to minimize \(S\).

\[S(x)=4x\left(\dfrac{216}{x^2}\right)+x^2.\nonumber \]

Therefore, \(S(x)=\dfrac{864}{x}+x^2\).

\[S′(x)=−\dfrac{864}{x^2}+2x.\nonumber \]

\[S(6\sqrt[3]{2})=\dfrac{864}{6\sqrt[3]{2}}+(6\sqrt[3]{2})^2=108\sqrt[3]{4}\,\text{in}^2\nonumber \]

## Exercise \(\PageIndex{6}\)

If the cost of one of the sides is \(30¢/\text{in}^2,\) the cost of that side is \(0.30xy\) dollars.

\(c(x)=\dfrac{259.2}{x}+0.2x^2\) dollars

## Key Concepts

- To solve an optimization problem, begin by drawing a picture and introducing variables.
- Find an equation relating the variables.
- Find a function of one variable to describe the quantity that is to be minimized or maximized.
- Look for critical points to locate local extrema.

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## AP®︎/College Calculus AB

## Optimization: box volume (Part 1)

- Optimization: box volume (Part 2)
- Optimization: profit
- Optimization: cost of materials
- Optimization: area of triangle & square (Part 1)
- Optimization: area of triangle & square (Part 2)
- Motion problems: finding the maximum acceleration

## Want to join the conversation?

## Video transcript

## RadfordMathematics.com

- maximizing : means finding the largest (or maximum) value the quantity can be
- minimizing : means finding the smallest (or minimum) value the quantity can be

## Must-Know Optimization Problem 1

To construct a rectangular enclosure for his chickens, Paul uses 40m of fencing.

What dimensions should his enclosure be for the area to be maximized?

## Must-Know Optimization Problem 1 - Answer Key

## Must-Know Optimization Problem 1 - Detailed Solution

We start by writing the integral in terms of \(t\) .

- Step 1: The quantity we're trying to optimize (maximize in this case) is the rectangular surface area of Paul's enclosure. The variables are the width \(x\) and the "height" \(y\) of the rectangle making the enclosure. The enclosed area is therefore: \[A = xy\]
- Step 2: The constraint is the fact that Paul only has 40m of fencing. Those 40m are linked to the variables \(x\) and \(y\) by the fact that: \[40 = 2x+y\]
- Step 3: rearranging \(40 = 2x + y\) we can make \(y\) the subject: \[y = 40 - 2x\] Now we can replace \(y\) in our expression for the area, \(A=xy\), to express \(A\) in terms of \(x\) only: \[A = x\begin{pmatrix}40 - 2x\end{pmatrix}\] that's: \[A = 40x - 2x^2\] which we can write as a function of \(x\): \[A(x) = 40x - 2x^2\]
- Step 4: to find the maximum area, we turn to calculus and start by differentiating \(A(x)\): \[A'(x)=40 - 4x\] We now solve \(A'(x)=0\), that's: \[\begin{aligned} & 40 - 4x = 0 \\ & 40 = 4x \\ & 10 = x \\ & x = 10 \end{aligned}\] To check that \(x=10\) corresponds to a maximum, we can use the second derivative test : \[A''(x) = -4\] So when \(x = 10\): \[A''(10) = -4 < 0 \] Since the second derivative is negative, we can confirm: \(x=10\) corresponds to the maximum. Using the expression found at the beginning of step 3 we can therefore state that: \[\begin{aligned} y & = 40 - 2\times 10 \\ & = 40 - 20 \\ y & = 20 \end{aligned}\]
- width : \(x = 10\)
- "height" : \(y= 20\)

## Must-Know Optimization Problem 1 - Video Solution

## Keep in Mind

Every optimization problem we'll come across can be solved used the four step method shown here.

## Tests & Exams

## Four Step Method for Solving Optimization Problems

Note: making a sketch for this step is often very useful.

## Must-Know Optimization Problem 2

## Must-Know Optimization Problem 2 - Answer Key

- radius: \(r = \sqrt[3]{\frac{165}{\pi}} \approx 3.74\)cm
- height: \(h = \frac{330}{\pi \begin{pmatrix} \sqrt[3]{\frac{165}{ \pi}}\end{pmatrix}^2} \approx 7.49\)cm

## Must-Know Optimization Problem 2 - Detailed Solution

- the sum of two discs, one at the top, the other at the bottom: \(\pi r^2 + \pi r^2 = 2\pi r^2\)
- the curved surface area of the can: \(2\pi r h\)

- Step 3 (use the constraint to eliminate one of the variables from the quantity we're optimizing): rearranging our expression for the constraint: \[h.\pi r^2 = 330\] we can make \(h\) the subject: \[h = \frac{330}{\pi r^2}\] Now we can replace \(h\) in our expression for the soda can's surface area, \(S = 2\pi r^2 + 2\pi r h\), to express \(S\) in terms of \(r\) only: \[\begin{aligned} S & = 2\pi r^2 + 2\pi r \times \frac{330}{\pi r^2} \\ & = 2 \pi r^2 + \frac{{660}\pi r}{\pi r^2} \\ S & = 2\pi r^2 + \frac{660}{r} \end{aligned}\] which we can write as a function of \(r\): \[S(r) = 2\pi r^2 + \frac{660}{r}\]
- Step 4 (use calculus to optimize): to find the can's minimum surface area, we turn to calculus and start by differentiating \(S(r)\) with respect to \(r\): \[S'(r) = 4\pi r - \frac{660}{r^2}\] We now solve \(S'(r)= 0\), that's: \[\begin{aligned} & 4\pi r - \frac{660}{r^2} = 0 \\ & 4 \pi r = \frac{660}{r^2} \\ & 4 \pi r^3 = 660 \\ & r^3 = \frac{660}{4\pi} = \frac{165}{\pi} \\ & r = \sqrt[3]{\frac{165}{\pi}} \approx 3.74 \ \text{cm} \end{aligned}\]

## Must-Know Optimization Problem 2 - Video Solution

## Must-Know Optimization Problem 3

## Must-Know Optimization Problem 3 - Answer Key

## Must-Know Optimization Problem 3 - Detailed Solution

- Step 1 (find an expression for the quantity we need to optimize, in term of the problem's variables): The quantity we're tyring to opyimize (minimize in this case) is the distance from the point \(\begin{pmatrix}2,0\end{pmatrix}\) to the curve \(y=\sqrt{x}\), which we can call \(D\); another way of thinking of this problem is that we're looking for the point on the curve \(y=\sqrt{x}\) that is closest to the point \(\begin{pmatrix}2,0\end{pmatrix}\). The distance from the point \(\begin{pmatrix}2,0\end{pmatrix}\) to any point \(\begin{pmatrix}x,y\end{pmatrix}\) in the \(xy\)-plane (whether that point be on the curve \(y=\sqrt{x}\) or not) is: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + \begin{pmatrix}y-0\end{pmatrix}^2}\] That's: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + y^2}\] Note : The formula we used was the formula for the distance beetween 2 points . It states that the distance, \(D\), between 2 points \(P\begin{pmatrix}x_1,y_1\end{pmatrix}\) and \(Q\begin{pmatrix}x_2,y_2\end{pmatrix}\) is: \[D = \sqrt{\begin{pmatrix}x_2-x_1\end{pmatrix}^2 + \begin{pmatrix}y_2-y_1\end{pmatrix}^2}\]
- Step 2 (write the constraint in terms of the variables): in this case the only thing stopping us from making the distance from the point \(\begin{pmatrix}2,0\end{pmatrix}\) equal to zero is the fact that the other point has to lie along the curve \(y=\sqrt{x}\) ; that is the constraint (the fact that \(y\) must equal to \(\sqrt{x}\)). So, the constraint is "simply": \[y=\sqrt{x}\]
- Step 3 (use the constraint to eliminate one of the variables from the quantity we're optimizing): Our expressing for the constraint is: \[y=\sqrt{x}\] Replacing the \(y\) inside our expression for the distance, \(D=\sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + y^2}\), we obtain: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + \begin{pmatrix}\sqrt{x}\end{pmatrix}^2}\] That's: \[D = \sqrt{\begin{pmatrix}x-2\end{pmatrix}^2 + x}\] Expanding \(\begin{pmatrix}x-2\end{pmatrix}^2\): \[D = \sqrt{x^2 - 4x + 4 + x}\] gathering like terms we find: \[D(x) = \sqrt{x^2 - 3x + 4}\] We can now see that the distance is only a function of \(x\), \(D(x)\), and that we have successfully eliminated the variable \(y\).

## Must-Know Optimization Problem 3 - Video Solution

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