5 6 solving optimization problems homework answer key

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5.10 Introduction to Optimization Problems

1 min read • june 18, 2024

We previously learned how to find the minimum or maximum value of a function on an interval using either the First Derivative Test or the Second Derivative Test. What significance does figuring out these pieces of information have? Well, they can help us solve optimization problems!

👍 Optimization Problems

You may ask, what are optimization problems ? If we think about what the word “optimizing” means, we see that these types of problems involve finding the best possible solution, often looking to maximize or minimize a certain quantity.

So how can we use calculus to solve these kinds of problems? Well, if we go back to its definition, we see that we want to “look to maximize or minimize a certain quantity.” Do we know how to find the minimum or maximum value of a function? We certainly do! How? By taking the derivatives of the function (applying either the First Derivative or Second Derivative Tests )!

One tricky part about optimization problems is that they usually add one more variable that we have to account for. But don’t fear, this just means you have to do the extra step of finding a relationship between two of the variables so that you can “get rid of” one through substitution.

✏️ Optimization Walkthrough

Let’s go through a problem below to get a better idea of how to solve this type of problem:

Let s = x 2 + y 2 s = x^2 + y^2 s = x 2 + y 2 . If x y = − 36 xy = -36 x y = − 36 , what x x x and y y y values minimize s s s ?

Since there are two variables, x x x and y y y , and we do not know how to differentiate with respect to more than one variable, we should look to re-write one of them in terms of the other. This way we can differentiate with respect to just one variable, which we’ve done many times before!

Because it is given that x y = − 36 xy = -36 x y = − 36 , we can re-write y y y to be − 36 x -\frac{36}{x} − x 36 ​ .

So we have,

Note: s s s becomes s ( x ) s(x) s ( x ) because we have re-written the function of s s s rewritten in terms of x x x

Now, we just have to find the minimum of this function!

To do so, let’s apply the Second Derivative Test.

First, we need to find the critical points of the function. Since we know that critical points are when s ′ ( x ) = 0 s'(x)=0 s ′ ( x ) = 0 :

Next, we need to determine the concavity of s ( x ) s(x) s ( x ) at these points to verify they are minimum(s).

Since the power of x x x is even, s ′ ′ ( x ) s''(x) s ′′ ( x ) is positive for any x x x . So, s ( x ) s(x) s ( x ) is concave up at both x = − 6 x=-6 x = − 6 and 6 6 6 .

Therefore, both x = − 6 x=-6 x = − 6 and 6 6 6 will minimize s ( x ) . s(x). s ( x ) . If x = − 6 x=-6 x = − 6 , based on the given product of x x x and y y y , y = 6 y=6 y = 6 . Similarly, if x = 6 x=6 x = 6 , y = − 6 y=-6 y = − 6 .

The final solution is ( − 6 , 6 ) (-6, 6) ( − 6 , 6 ) and ( 6 , − 6 ) (6, -6) ( 6 , − 6 ) . Great work!

📝 Optimization Practice Problems

Now, let’s do some practice on your own!

❓ Optimization Problems

Optimization question 1.

An open-topped play area with a square base is designed to hold 32 32 32  cubic feet of sand. What is the minimum exterior surface area of the play area?

121 121 121  square inches of text is to be printed on a card. If there are to be exactly one-inch margins around all four sides of the text, what is the width and height of the smallest card that can be used?

✅ Optimization Answers and Solutions

The answer to this question is 48 48 48 square feet. Here’s why:

First, let us set the side length of the square base to be x x x and the height of the play area to be h h h .

This means that the volume of the play area can be expressed as

Since it is given that the volume is 32 32 32 cubic feet,

Since we can only differentiate with respect to one variable, we can re-write h h h in terms of x x x ,

In this problem, we want to minimize the surface area of the open-topped play area, which we can express as

This further simplifies to

Now, we just have to find the minimum of this function! Let’s apply the Second Derivative Test.

First, we need to find the critical points of the function.

Next, we need to determine the concavity of S ( x ) S(x) S ( x ) at this point to verify it is a minimum.

Since 6 > 0 6>0 6 > 0 , S ( x ) S(x) S ( x ) is concave up at x = 4 x=4 x = 4 .

Therefore, by the Second Derivative Test, x = 4 x=4 x = 4 minimizes S ( x ) S(x) S ( x ) .

This makes the minimum surface area of the open-topped play area

Optimization Question 2

The answer to question 2 is 13 13 13 inches by 13 13 13 inches. Here’s the work!

If x x x is the width of the printed text and y y y is the height of the printed text because we are given that the area of the printed text is 121 121 121 square inches, we can re-write y y y as:

Using the information given about the one-inch margins on all four sides of the text, we can write the width of the card as:

And write the height of the card as:

This means that the area of the card is

This can be expanded and simplified into

First, we need to find the critical points of the function, which are the points where A ′ ( x ) = 0 A'(x)=0 A ′ ( x ) = 0 .

We want the solution that is positive because width cannot be negative.

To verify that x = 11 x=11 x = 11 is a minimum, we need to determine the concavity of A ( x ) A(x) A ( x ) at this point.

A ( x ) A(x) A ( x ) is concave up at x = 11 x=11 x = 11 .

Therefore, by the Second Derivative Test, x = 11 x=11 x = 11 minimizes S ( x ) S(x) S ( x ) .

This means that the width of the printed area is 11 11 11 inches and the height of the printed region is 11 11 11 inches as well.

Therefore, the dimensions of the card itself are 13 13 13 inches by 13 13 13 inches.

Great work so far! In the next study guide , we’ll do a couple more optimization problems to really get that practice in. You’re so close to finishing unit 5 of AP Calculus. 🥳

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Calculus Optimization with Lesson Video (Unit 5)

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Calculus Optimization - Solving Real-World Problems to Maximize or Minimize Lesson:

Your AP Calculus students will use techniques of differentiation to solve applied minimum and maximum problems. Your students will have guided notes, homework, and a content quiz on Optimization - Solving Real-World Problems to Maximize or Minimize that cover the concepts in depth from the six-lesson unit on Analytical Applications of Differentiation. #distancelearningtpt

***(1/04/23) EX #6 added the answer to the last question for the key.

What is included in this resource?

⭐ Guided Student Notes

⭐ Fully-editable SMART Board® Slides

⭐ Homework/Practice assignment

⭐ Content quiz

⭐ Video Lesson Link for Distance Learning - Flipped Classroom models

⭐ Full solution set

Lesson Objective:

★ Use calculus to optimize real-world situations by finding values that maximize or minimize given constraints.

******************************************************************************************************************************************************

NOTE: This lesson has been updated to reflect the changes in the AP Calculus CED Binder and includes a LESSON VIDEO for Distance Learning needs.

*** ***************************************************************************************************************************************************

College Board® Topics:

Topic 5.10: Introduction to Optimization Problems

Topic 5.11: Solving Optimization Problems

Topic 5.12: Exploring Behaviors of Implicit Relations

College Board® Learning Objectives:

FUN-4.B: Calculate minimum and maximum values in applied contexts or analysis of functions.

FUN-4.B.1: The derivative can be used to solve optimization problems; that is, finding a minimum or maximum value of a function on a given interval.

FUN-4.C: Interpret minimum and maximum values calculated in applied contexts.

FUN-4.C.1: Minimum and maximum values of a function take on specific meanings in applied contexts.

FUN-4.D: Determine critical points of implicit relations

FUN-4.D.1: A point on an implicit relation where the first derivative equals zero or does not exist is a critical point of the function.

FUN-4.E: Justify conclusions about the behavior of an implicitly defined function based on evidence from its derivatives

FUN-4.E.1: Applications of derivatives can be extended to implicitly defined functions.

FUN-4.E.2: Second derivatives involving implicit differentiation may be relations of x , y , and dy / dx .

The unit includes:

1) Mean Value Theorem

2) Extrema on an Interval

3) First Derivative Test

4) Second Derivative Test

5) Curve Sketching

6) Optimization

Click HERE to SAVE 20% by buying all ANALYTIC APPLICATIONS OF DIFFERENTIATION products, including cooperative activities, in UNIT 5 MEGA BUNDLE.

You might be interested in:

Applications of the Derivative Scavenger Hunt

Maximize-A-Box Optimization Project

Analytic Applications of Differentiation Unit Homework

Analytic Applications of Differentiation Assessments

Analytic Applications of Differentiation Essentials

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• Calculus Home

How to Solve Optimization Problems in Calculus

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

Need to solve Optimization problems in Calculus? Let’s break ’em down and develop a strategy that you can use to solve them routinely for yourself.

• Find the largest ….
• Find the minimum….
• What dimensions will give the greatest…?

The first stage doesn’t involve Calculus at all, while by contrast the second stage is just a max/min problem that you recently learned how to solve: Stage I. Develop the function. You must first convert the problem’s description of the situation into a function — crucially, a function that depends on only one single variable.

Stage II. Maximize or minimize that function. Now maximize or minimize the function you just developed. You’ll use your usual Calculus tools to find the critical points, determine whether each is a maximum or minimum, and so forth. We’ll break these two big Stages into smaller steps below. To illustrate those steps, let’s together solve this classic Optimization example problem:

Stage I. Develop the function

Step 2. Most frequently you’ll use your geometry knowledge. Having drawn the picture, the next step is to write an equation for the quantity we want to optimize . Most frequently you’ll use your everyday knowledge of geometry for this step. In this problem, for instance, we want to minimize the cost of constructing the can, which means we want to use as little metal as possible . Hence we want to minimize the can’s surface area . So let’s write an equation for that total surface area:

\begin{align*} A_\text{total} &= A_\text{top} + A_\text{cylinder} + A_\text{bottom} \\[8px] &= \pi r^2 + 2\pi r h + \pi r^2 \\[8px] &= 2\pi r^2 + 2 \pi r h \end{align*}

That’s it; you’re done with Step 2! You’ve written an equation for the quantity you want to minimize $(A_\text{total})$ in terms of the relevant quantities ( r and h ).

• Optimization Problems & Complete Solutions

Step 3. Here’s a key thing to know about how to solve Optimization problems: you’ll almost always have to use detailed information given in the problem to rewrite the equation you developed in Step 2 to be in terms of one single variable.

Above, for instance, our equation for $A_\text{total}$ has two variables, r and h . We must eliminate one of them in order to proceed. The choice of which to keep and which to eliminate is arbitrary; for our solution here, we choose to keep r . (We could just as easily choose h , and develop our solution along that path instead. We’d arrive at the same final result.) Since we’re choosing to work with r , we need to use other detailed information given in the problem to write h in terms of r so we can substitute for h as a variable.

Begin subproblem.

To accomplish this substitution, we look back to see what other constraints/information the problem gave us: recall that the can must hold an amount V of liquid, where V is some number. ( V might be 355 cm$^3$, for instance.) Now a cylinder of radius r and height h has a volume of $V = \pi r^2 h,$ and so we can solve for h in terms of V and constants: $$V = \pi r^2 h$$ thus $$h = \dfrac{V}{\pi r^2} $$ That’s our expression for h in terms of r (and the constants V and $\pi).$ End subproblem.

We’re done with Step 3: we now have the function in terms of a single variable, r : $$A(r) = 2\pi r^2 + \frac{2V}{r}$$ We’re now writing $A(r)$ to emphasize that A is a function of only the single variable r , and we’ve dropped the subscript “total” from $A_\text{total}$ since we no longer need it.

This also concludes Stage I of our work: in these threes steps, we’ve developed the function we’re now going to minimize!

Notice, by the way, that so far in our solution we haven’t used any Calculus at all. That will always be the case when you solve an Optimization problem: you don’t use Calculus until you come to Stage II.

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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:

You would probably automatically find the derivative $A'(r)$ (which you could equivalently write as $\dfrac{dA}{dr})$, then find the critical points, then determine whether each represents a maximum or a minimum for the function, and so forth. That’s exactly what we’re now going to do in Stage II. Hence, you already know how to do all of the following steps; the only new part to maximization problems is what we did in Stage I above.

Step 4. We want to minimize the function $$ A(r) = 2\pi r^2 + \frac{2V}{r}$$ and so of course we must take the derivative, and then find the critical points .

The critical points occur when $A'(r) = 0$: \[ \begin{align*} A'(r) = 0 &= 4 \pi r\, – \frac{2V}{r^2} \\[8px] \frac{2V}{r^2} &= 4 \pi r \\[8px] \frac{2V}{4 \pi} &= r^3 \\[8px] r^3 &= \frac{V}{2\pi} \\[8px] r &= \sqrt[3]{\frac{V}{2\pi}} \end{align*} \] We thus have only one critical point to examine, at $r = \sqrt[3]{\dfrac{V}{2\pi}}\,.$

Step 5. Next we must justify that the critical point we’ve found represents a minimum for the can’s surface area (as opposed to a maximum, or a saddle point). We could reason physically, or use the First Derivative Test, but we think it’s easiest in this case to use the Second Derivative Test. Let’s quickly compute the second derivative, starting with the first derivative that we found above:

Since $r > 0$, this second derivative $\left(A’^\prime(r) = 4\pi + \dfrac{4V}{r^3}\right)$ is always positive $\left(A’^\prime(r) > 0 \right)$. That is, the graph of A(r) versus r is always concave up. Hence this single critical point gives us a minimum (as opposed to a maximum or saddlepoint), which is what we’re after:

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

Step 6. Now that we’ve found the critical point that corresponds to the can’s minimum surface area (thereby minimizing the cost), let’s finish answering the question : The problem asked us to find the dimensions — the radius and height — of the least-expensive can. We’ve already found the relevant radius, $r = \sqrt[3]{\dfrac{V}{2\pi}}\,.$

To find the corresponding height, recall that in the Subproblem above we found that since the can must hold a volume V of liquid, its height is related to its radius according to $$h = \dfrac{V}{\pi r^2}\,. $$ Hence when $r = \sqrt[3]{\dfrac{V}{2\pi}}\,,$ \[ \begin{align*} h &= \frac{V}{\pi}\,\frac{1}{r^2} \\[8px] &= \frac{V}{\pi}\,\frac{1}{\left( \sqrt[3]{\frac{V}{2\pi}}\right)^2} \\[8px] &= \frac{V}{\pi}\,\frac{2^{2/3}\pi^{2/3}}{V^{2/3}} \\[8px] &= 2^{2/3}\frac{V^{1/3}}{\pi^{1/3}} \\[8px] h &= 2^{2/3}\sqrt[3]{\frac{V}{\pi}} \quad \triangleleft \end{align*} \] The preceding expression for h is correct, but we can gain a nice insight by noticing that $$2^{2/3} = 2 \cdot\frac{1}{2^{1/3}}$$ and so \[ \begin{align*} h &= 2^{2/3}\sqrt[3]{\frac{V}{\pi}} \\[8px] &= 2 \cdot\frac{1}{2^{1/3}}\,\sqrt[3]{\frac{V}{\pi}} \\[8px] &= 2 \sqrt[3]{\frac{V}{2\pi}} = 2r \end{align*} \] since recall that the ideal radius is $r = \sqrt[3]{\dfrac{V}{2\pi}}\,.$ Hence the ideal height h is exactly twice the ideal radius.

Step 7. One last check You’ll lose points if you don’t answer the question that was asked. Because Optimization solutions can be long, we recommend that before finishing you go back and check what quantity/quantities the problem requested, and make sure you’ve provided that — especially on an exam, where you’ll lose points if you don’t answer the exact question that was asked. For example, the problem could have asked to find the value of the smallest possible surface area A , or the minimum cost.

Instead, in this case, the problem stated, “What dimensions (height and radius) will minimize the cost of metal to construct the can?” We have provided those two dimensions, and so we are done. $\checkmark$

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Summary: Problem Solving Strategy

We’ve now illustrated the steps we use to solve every single Optimization problem we encounter, and they always work. PROBLEM SOLVING STRATEGY: Optimization The strategy consists of two Big Stages. The first does not involve Calculus at all; the second is identical to what you did for max/min problems.

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.

For now, over to you:

• 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 !
• How can we make posts such as this one more useful to you?

Please head to our Forum and post!

[Thanks to S. Campbell for his specific research into students’ learning of Optimization:

“College Student Difficulties with Applied Optimization Problems in Introductory Calculus,” unpublished masters thesis, The University of Maine, 2013.]

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What are your thoughts or questions.

What a phenomenal explanation, you just saved my ap calculus grade haha

We’re happy to have helped. Good luck with your exam! : )

Man, if only everyone was a thorough in their explanations as this one is

Thank you for the compliment. We are very happy to have helped! : )

Nice explanation and methodical approach. Setting up the problem is 99% of the problem. I’m still trying to figure out on other optimization solutions what yo do if the 2nd derivative is simply a constant. If f ‘(x) has a desired min or max, and f ‘’(x) differentiates to a constant, what does this mean to the max or min in the first derivative test? Does the sign of the constant alone in f ‘’(x) then determine concavity since there’s no potential inflection point? If that constant’s sign is negative (concave down) and the solution requires an optimization max, does that satisfy a proof? And vice versa if concavity is positive (concave up) confirm a minimum in the first derivative test?

Thanks, Chris. We’re glad to know you liked our explanation and approach. And agreed about getting the problem set-up right as the vast majority of the work here.

The answer to all of your questions is: yes! If the second derivative is a negative constant, then the function is concave down everywhere, and so you’re guaranteed that the point x=c you found where f'(c) = 0 is a maximum. (See the figure below.) Similarly, if the second derivative is a positive constant, then the function is concave up everywhere, and so the point x=c where f'(c) = 0 is guaranteed to be a minimum. And the fact that there’s no point of inflection anywhere doesn’t affect those conclusions.

The only thing that you wrote that isn’t quite right are the very last words, “in the first derivative test”; instead, you’re using the Second Derivative Test . That test is just as conclusive as the First Derivative Test, and is often easier to use. The one exception is if the second derivative is zero at the point of interest (f”(c)=0), in which case the Second Derivative Test is inconclusive and you have to revert to the First Derivative Test. But otherwise, the conclusion you reach with the Second Derivative test is indeed conclusive.

Hope that helps, and thanks for asking!

what problems can help to solve optimization

Thanks for asking! We have more completely solved optimization problems on this page: Optimization: Problems and Solutions .

We hope that helps!

very nicely organized! however i think it would have been more effective with some numbers, instead of variables. it can get hard to follow, especially when there’s multiple(in this case). but it was still lovely and easy to follow

Thank you for your nice comment, and for your suggestion. We’ll keep it in mind for future posts. For now: thanks very much!

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Section 5.6 - Optimization and Modeling

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