Optimization

This page contains free practice, typical optimization problems, each with a detailed solution. The only way to get this problem type down is by practicing, so practice here where you won't be penalized for getting stuck or making small errors. You are guaranteed to have an optimization problem on your upcoming exam, so the time to make all those typical errors is now!

We work through how to use our "Problem Solving Strategy: Optimization" step-by-step in our blog post How to Solve Optimization Problems in Calculus, one of our most popular posts.

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.

Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only one variable. The steps:

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.

You now have a standard max/min problem to solve.

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.

Minimize cost of materials for a soda can

Practice Problem: Least expensive can

A cylindrical can, with no top lid, must contain

𝑉

3

of liquid. What dimensions will minimize the cost of metal to construct the can?

Stage I: Develop the function.

Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only one variable. The steps:

$A cylindrical open-topped can has height h and radius r, and is open at the top$ 1. Draw a picture of the physical situation.
See the figure. We've called the radius of the cylinder r, and its height h.

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
We want to minimize the amount of metal we use, which is to say we want to minimize the area of the can. The can consists of a cylinder of surface area

𝐴 c y l i n d e r = (2 𝜋 𝑟) (ℎ)

, and a bottom piece that has area

𝐴 b o t t o m = 𝜋 𝑟 2

. The can has no top. Its total area is thus:

𝐴 = 𝐴 c y l i n d e r + 𝐴 b o t t o m = 2 𝜋 𝑟 ℎ + 𝜋 𝑟 2

3. If necessary, use other given information to rewrite your equation in terms of a single variable.
The can's area A currently depends on two variables, r and h. In order to proceed, we must use other information we're given to rewrite the area in terms of just one of those variables. Let's choose r as that single variable. We must then eliminate the height h as a variable. To do so, we note that the can must hold volume V of liquid. Since the volume of a cylinder of radius r and height h is

𝑉 = 𝜋 𝑟 2 ℎ

We can therefore solve for h:

ℎ = 𝑉 𝜋 𝑟 2

Substituting this expression for h into the expression for the can's surface area:

𝐴 = 2 𝜋 𝑟 ℎ + 𝜋 𝑟 2 = 2 𝜋 𝑟 (𝑉 𝜋 𝑟 2) + 𝜋 𝑟 2 = 2 𝜋 𝑟 (𝑉 𝜋 𝑟 2) + 𝜋 𝑟 2 = 2 𝑉 𝑟 + 𝜋 𝑟 2

The expression for A is now a function of the single variable r. (Remember that V is a constant that just tells us how much liquid the can must hold.) graph of the can's area as a function of its radius

graph of the can's area as a function of its radius

We've graphed the function (with

𝑉 = 355 c m 3

), a step you probably wouldn't do yourself — but we want to emphasize that everything you've done so far is to create a function that you're now going to minimize. One choice would be to closely examine the graph to determine the value of r that minimizes A . . . but instead we're going to use the max/min techniques you learned recently!

Stage II: Maximize or minimize the function.

You now have a standard max/min problem to solve. 4. Take the derivative of your equation with respect to your single variable. Then find the critical points.

𝑑 𝐴 𝑑 𝑟 = 𝑑 𝑑 𝑟 [2 𝑉 𝑟 + 𝜋 𝑟 2] = - 2 𝑉 𝑟 2 + 2 𝜋 𝑟

The critical points occur when

𝑑 𝐴 𝑑 𝑟 = 0

𝑑 𝐴 𝑑 𝑟 = 0 = - 2 𝑉 𝑟 2 + 2 𝜋 𝑟 2 𝑉 𝑟 2 = 2 𝜋 𝑟 𝑟 3 = 𝑉 𝜋 𝑟 = 3 \sqrt 𝑉 𝜋

5. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.
Let's examine the second derivative. Above we found the first derivative:

𝑑 𝐴 𝑑 𝑟 = - 2 𝑉 𝑟 2 + 2 𝜋 𝑟

𝑑 2 𝐴 𝑑 𝑟 2 = 𝑑 𝑑 𝑟 (- 2 𝑉 𝑟 2 + 2 𝜋 𝑟) = - 2 (- 2 𝑉 𝑟 3) + 2 𝜋 = 4 𝑉 𝑟 3 + 2 𝜋

Since

𝑟 > 0

, this second derivative

(𝑑 2 𝐴 𝑑 𝑟 2 = 4 𝑉 𝑟 3 + 2 𝜋)

is always positive

(𝑑 2 𝐴 𝑑 𝑟 2 > 0)

. Hence this single critical point gives us a minimum:

The minimum area occurs when $𝑟 = 3 \sqrt 𝑉 𝜋$ .

6. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
Recall that we found above

ℎ = 𝑉 𝜋 𝑟 2

. Hence when

𝑟 = 3 \sqrt 𝑉 𝜋

, the cylinder's height is

ℎ = 𝑉 𝜋 𝑟 2 = 𝑉 𝜋 (3 \sqrt 𝑉 𝜋) 2 = 𝑉 𝜋 𝜋 2 / 3 𝑉 2 / 3 = 3 \sqrt 𝑉 𝜋

Hence we obtain the minimum area for this open-topped can when the cylinder's height equals its radius, and the dimensions are:

ℎ = 𝑟 = 3 \sqrt 𝑉 𝜋 ✓

7. Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .
The question asked us to specify the cylinder's dimensions, which we have provided.

✓

[close solution]

What are the dimensions of the poster with the smallest total area?

Practice Problem: Printed Poster

A poster must have a printed area of 320 cm

2

. It will have top and bottom margins that are 5 cm each, and side margins that are 4 cm. What are the dimensions of the poster with the smallest total area?

Answer: Width = 24 cm, and length =30 cm

Stage I: Develop the function.

Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only one variable. The steps: $As described in text$ 1. Draw a picture of the physical situation.
See the figure. We've called the width of the printed area x, and its length y. We can then write the printed area as

𝐴 p r i n t e d = 𝑥 𝑦 = 320 c m 2

Note that this picture captures the key features of the situation, and we wouldn't make good progress without it!

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
We see from the figure that the poster has total width

𝑤 = 𝑥 + 8

cm, and total length

𝑙 = 𝑦 + 10

cm. Its total area is thus

𝐴 t o t a l = (𝑥 + 8) (𝑦 + 10)

3. If necessary, use other given information to rewrite your equation in terms of a single variable.
The poster's area

𝐴 t o t a l

currently depends on two variables, x and y. In order to proceed, we must use other information we're given to rewrite the area in terms of just one of those variables. Let's choose the width x as that single variable. We must then eliminate the height y as a variable. To do so, recall that the printed area is given by

𝐴 p r i n t e d = 𝑥 𝑦 = 320 c m 2

Hence we can write y in terms of x as

𝑦 = 320 𝑥

Substituting this expression for y into our expression for the poster's total area

𝐴 t o t a l

gives:

𝐴 t o t a l = (𝑥 + 8) (𝑦 + 10) = (𝑥 + 8) (320 𝑥 + 10) = 10 (𝑥 + 8) (32 𝑥 + 1) = 10 (32 + 𝑥 + 256 𝑥 + 8) = 10 (𝑥 + 256 𝑥 + 40)

Poster's area as a function of its width x

We've graphed the function, a step you probably wouldn't do yourself — but we want to emphasize that everything you've done so far is to create a function that you're now going to minimize. One choice would be to closely examine the graph to determine the value of x that minimizes A . . . but instead we're going to use the max/min techniques you learned recently!

Stage II: Maximize or minimize the function.

You now have a standard max/min problem to solve.

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.

𝑑 𝐴 t o t a l 𝑑 𝑥 = 𝑑 𝑑 𝑥 [10 (𝑥 + 256 𝑥 + 40)] = 10 𝑑 𝑑 𝑥 [(𝑥 + 256 𝑥 + 40)] = 10 [1 - 256 𝑥 2]

The critical point occurs when

𝑑 𝐴 t o t a l 𝑑 𝑥 = 0

𝑑 𝐴 t o t a l 𝑑 𝑥 = 0 = 10 [1 - 256 𝑥 2] 0 = 1 - 256 𝑥 2 𝑥 2 = 256 𝑥 = \sqrt 256 = 16 c m

5. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
Recall that we found above that

𝑦 = 320 𝑥

. Hence hen the printed area's width is

𝑥 = 16

cm, its height is

𝑦 = 32016 = 20 c m

6. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.
Let's use the second derivative test to verify that the critical point we've found represents a minimum. We found above that the first derivative is

𝑑 𝐴 t o t a l 𝑑 𝑥 = 10 (1 - 256 𝑥 2)

The second derivative is thus

𝑑 2 𝐴 t o t a l 𝑑 𝑥 2 = 10 𝑑 𝑑 𝑥 (1) - 10 (256) 𝑑 𝑑 𝑥 (1 𝑥 2) = 0 - (- 2) (10) (256) 1 𝑥 3 = (2) (10) (256) 𝑥 3

Since the width is always positive (

𝑥 > 0

), this second derivative is always positive, and so our single critical point gives us a minimum value for the poster's area.

7. Finally, check to make sure you have answered the question as asked: x or y values, or coordinates, or a maximum area, or a shortest time, or . . . .
Careful! We've found x and y for the poster's printed area, but the question asks for the dimensions of the overall poster. Those values are

w i d t h = 𝑥 + 8 = 16 + 8 = 24 c m l e n g t h = 𝑦 + 10 = 20 + 10 = 30 c m

The minimum dimensions of the poster are thus width

𝑥 = 24

cm and length

𝑦 = 30

cm.

✓

[close solution]

Shortest time to row, then run, to a point on shore

Practice Problem: Shortest time to row, then run, to a point on shore

$As described in text.$ You are in a boat on a still lake, near a straight section of shore. The nearest point to shore is 2 km away. You are aiming to get back to your food and drink supplies, 4 km south of that nearest point. You can row at 5 km/hr, and run at 8 km/hr. At what point

𝑃

should you land your boat to reach your supplies as quickly as possible---before they are consumed by faster friends?

Answer: 1.6 km south

$As described in text$ 1. Draw a picture of the physical situation.
See the figure. We've called the distance from the point nearest on shore down to where will land the boat

𝑥

.

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
We want to minimize the time it will take us to reach the supplies. Recall the fundamental relation

𝑡 𝑖 𝑚 𝑒 = 𝑑 𝑖 𝑠 𝑡 𝑎 𝑛 𝑐 𝑒 𝑟 𝑎 𝑡 𝑒

For instance, if we were to row straight to shore at 5 km/hr, traversing 2 km, that would take

2 k m / h r 5 k m / h r = 0.40

hr. And then running south the 4 km at 8 km/hr would take

4 k m / h r 8 k m / h r = 0.50

hr, for a total travel time of 0.9 hours. We can shorten this time by doing more rowing and landing at point

𝑃

further south.

Or you could instead row straight to your supplies, which would take time

\sqrt 22 + 42 5 = 0.89

hr.

Let's use our usual approach to find a minimum time required if we put the point

𝑃

somewhere between those two extremes.

By landing at point

𝑃

, we row a distance (by the Pythagorean Theorem)

r o w i n g d i s t a n c e = \sqrt 22 + 𝑥 2 = \sqrt 4 + 𝑥 2

Hence the time required to row, at 5 km/hr, is

r o w i n g t i m e = r o w i n g d i s t a n c e r o w i n g r a t e = \sqrt 4 + 𝑥 2 5

We then run a distance

(4 - 𝑥)

km, at a rate of 8 km/hr, thus taking time

r u n n i n g t i m e = r u n n i n g d i s t a n c e r u n n i n g r a t e = 4 - 𝑥 8

The total time is thus given by the function

𝑓 (𝑥)

𝑓 (𝑥) = \sqrt 4 + 𝑥 2 5 + 4 - 𝑥 8

Since our equation has only one variable, x, we skip step 3 of our strategy.

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.

𝑑 𝑓 𝑑 𝑥 = 𝑑 𝑑 𝑥 [\sqrt 4 + 𝑥 2 5 + 4 - 𝑥 8] = 12 1 5 \sqrt 4 + 𝑥 2 (2 𝑥) - 18 = 𝑥 5 \sqrt 4 + 𝑥 2 - 18

The minimum time occurs when

𝑑 𝑓 𝑑 𝑥 = 0

𝑑 𝑓 𝑑 𝑥 = 0 = 𝑥 5 \sqrt 4 + 𝑥 2 - 18 𝑥 5 \sqrt 4 + 𝑥 2 = 18 8 𝑥 = 5 \sqrt 4 + 𝑥 2 64 𝑥 2 = 25 (4 + 𝑥 2) 64 𝑥 2 = 100 + 25 𝑥 2 39 𝑥 2 = 100 𝑥 = \sqrt 10039 = 1.6 k m

5. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
6. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.
Just for completeness, let's compute the time required when you land 1.6 km south of your current location:

t o t a l t i m e = \sqrt 4 + 𝑥 2 5 + 4 - 𝑥 8 = \sqrt 4 + (1.6) 2 5 + 4 - 1.6 8 = 0.81 h r

This time is less than the two times we computed earlier (0.9 hr if you row straight to shore; 0.89 hr if you row straight to your supplies).

7. Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .
The best location to land is a distance

𝑥 = 1.6

km south, in order to minimize the time required to reach your supplies.

✓

[close solution]

Two boats sailing; when closest together?

Practice Problem: Sailing ships

Boat A leaves a dock at noon and travels due north at 15 km/hr. Simultaneously, Boat B is heading due east toward the same dock at 20 km/hr and reaches there at 2:00 PM. At what time are the two boats closest together?

Answer: 1.28 hr

$As described in text$ 1. Draw a picture of the physical situation.
Once again, a good picture is key to proceeding correctly. Here we'll actually sketch two figures to represent the situation at noon, when the scenario begins, and one for some later, general time

𝑡

.

The upper figure shows the situation at noon: Boat A is leaving the dock (shown as a brown square), traveling northward. Boat B is headed straight eastward toward the dock. At this instant, it is 40 km away, which we know since it will reach the dock 2 hours later (at 2:00 PM) and is traveling at 20 km/hr.

The lower figure shows the situation at later time

𝑡

, where

𝑡

is measured in hours. Since Boat A is traveling at 15 km/hr, it has traveled northward a distance

𝑦 = 15 𝑡

northward of the dock.

Since Boat B is traveling at 20 km/hr, it has traveled a distance

20 𝑡

, meaning it is now a distance

𝑥 = 40 - 20 𝑡

away from the dock.

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
The distance

𝑑

between the boats at time

𝑡

is given by the Pythagorean theorem:

𝑑 2 = 𝑥 2 + 𝑦 2 = (40 - 20 𝑡) 2 + (15 𝑡) 2

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.
The minimum value of

𝑑

occurs at the same instant as when

𝑑 2

is minimized, so let's define the function

𝑓 (𝑡) = 𝑑 2 = (40 - 20 𝑡) 2 + (15 𝑡) 2

and minimize

𝑓

. (Otherwise we would write

𝑑

as the square root of everything on the right-hand side of the equation, but that just introduces a messy square root we then have to work with.)

𝑑 𝑓 𝑑 𝑡 = 𝑑 𝑑 𝑡 [(40 - 20 𝑡) 2 + (15 𝑡) 2] = 2 (40 - 20 𝑡) (- 20) + 2 (15 𝑡) (15) = (40 - 20 𝑡) (- 40) + (30 𝑡) (15) = 40 (- 40) - 20 𝑡 (- 40) + (30) (15 𝑡) = - 1600 + 800 𝑡 + 450 𝑡 = - 1600 + 1250 𝑡

Hence the critical point occurs for

𝑡

such that

𝑑 𝑓 𝑑 𝑡 = 0 = - 1600 + 1250 𝑡 1250 𝑡 = 1600 𝑡 = 16001250 = 1.28 h r ✓

𝑓

is positive (

𝑓 ″ (𝑡) = 1250

), and so we know this single critical point is gives the absolute minimum.

7. Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .
We have found the time at which the boats are closest, as requested.

✓

[close solution]

What dimensions minimize the cost of a garden fence?

Garden fence dimensions

Sam wants to build a garden fence to protect a rectangular 400 square-foot planting area. His next-door neighbor agrees to pay for half of the fence that borders her property; Sam will pay the rest of the cost. What are the dimensions of the planting area that will minimize Sam's cost to build the fence? (You may leave your answer as a square root; you don't have to find a decimal result.)

View/Hide Solution

1. Draw a picture of the physical situation.

See the figure. We've called the width of the garden x (the top and bottom portions of the fence), and the length of the garden y (the left and right sides). Note also that the total area of Sam's garden must be $𝐴 = 400 f t 2$ .

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.

We want to minimize Sam's cost for building the fence, which is the same as minimizing the amount of fence that he's paying for. Let's call that cost C. Remember that Sam is only paying for half of the cost of the right side of the fence, since his neighbor is paying for that other half. Then

𝐶 = (l e n g t h o f l e f t s i d e) + (l e n g t h o f t o p) + 12 (l e n g t h o f r i g h t s i d e) + (l e n g t h o f b o t t o m) = 𝑦 + 𝑥 + 12 𝑦 + 𝑥 = 32 𝑦 + 2 𝑥

3. If necessary, use other given information to rewrite your equation in terms of a single variable.

The cost C currently depends on two variables, y and x. In order to proceed, we must use other information we're given to rewrite the area in terms of just one of those variables. Let's choose x as that single variable.

We must then eliminate y as a variable. To do so, recall that Sam's garden must have area $𝐴 = 400 f t 2$ . Since the garden's rectangular area is given by
$𝐴 = 𝑥 𝑦 = 400$
we can solve for y in terms of x:
$𝑦 = 400 𝑥$
Substituting this expression for y into our expression above for the cost C:
$𝐶 = 32 𝑦 + 2 𝑥 = 32 (400 𝑥) + 2 𝑥 = 600 𝑥 + 2 𝑥$
The expression for C is now a function of the single variable x, as required.

Graph of garden fence cost versus width x

We've graphed the function, a step you probably wouldn't do yourself — but we want to emphasize that everything you've done so far is to create a function that you're now going to minimize.

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.

𝐶 = 600 𝑥 + 2 𝑥 = 600 𝑥 - 1 + 2 𝑥 𝑑 𝐶 𝑑 𝑥 = 𝑑 𝑑 𝑥 (600 𝑥 - 1) + 𝑑 𝑑 𝑥 (2 𝑥) = - 600 𝑥 - 2 + 2 = - 600 𝑥 2 + 2

The critical points occur when $𝑑 𝐶 𝑑 𝑥 = 0$ :
$𝑑 𝐶 𝑑 𝑥 = 0 = - 600 𝑥 2 + 2 - 2 = - 600 𝑥 2 𝑥 2 = 6002 = 300 𝑥 = \sqrt 300$
Note that we choose the positive square root since the width x cannot be negative. Also note that we could have a critical point where $𝑑 𝐶 𝑑 𝑥 = - 600 𝑥 2 + 2$ is undefined, which occurs when $𝑥 = 0$ . That answer makes no physical sense, though, since then Sam's garden would have zero area. We thus continue our analysis with the single critical point
$𝑥 = \sqrt 300$

5. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.

Let's examine the second derivative. Above we found the first derivative:
$𝑑 𝐶 𝑑 𝑥 = - 600 𝑥 - 2 + 2$

The second derivative is thus
$𝑑 2 𝐶 𝑑 𝑥 2 = 𝑑 𝑑 𝑥 (- 600 𝑥 - 2) + 𝑑 𝑑 𝑥 (2) = (- 2) (- 600) 𝑥 - 3 + 0 = 1200 𝑥 3$
Since $𝑥 > 0$ , this second derivative $(𝑑 2 𝐶 𝑑 𝑥 2 = 1200 𝑥 3)$ is always positive $(𝑑 2 𝐶 𝑑 𝑥 2 > 0)$ . Hence this single critical point gives us a minimum for the cost C:

The minimum cost occurs when $𝑥 = \sqrt 300$ ft.

6. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.

Recall that we found above that $𝑦 = 400 𝑥$ . Hence when the width $𝑥 = \sqrt 300$ , the garden's length y must be:
$𝑦 = 400 𝑥 = 400 \sqrt 300 = 400 \sqrt 100 \sqrt 3 = 400 10 \sqrt 3 = 40 \sqrt 3$
Hence Sam's cost is minimized when the garden has

width $𝑥 = \sqrt 300 f t, ✓$

and

length $𝑦 = 40 \sqrt 3 f t . ✓$

7. Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .

The question asked us to specify the garden's dimensions, which we have provided. $✓$

[close solution]

Rectangle inscribed in ellipse

Practice Problem: Rectangle inscribed in ellipse

What is the area of the largest rectangle that can be inscribed in the ellipse

𝑥 2 + 4 𝑦 2 = 16

? (Do not use a calculator; you may leave square roots in your answer.)

Answer: 16

$As described in text.$ 1. Draw a picture of the physical situation.
See the figure. We've marked the upper right corner of the rectangle with coordinates

(𝑥, 𝑦)

, where

𝑥

and

𝑦

are related by the ellipse's equation since the rectangle's corners lie on the ellipse. The rectangle then has width

𝑤 = 2 𝑥

, and height

ℎ = 2 𝑦

.

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
We want to maximize the rectangle's area, given the constraint that its corners must lie on the ellipse.

𝐴 = (2 𝑥) (2 𝑦)

3. If necessary, use other given information to rewrite your equation in terms of a single variable.
We choose to write the ellipse's area solely in terms of

𝑥

, and so we must eliminate

𝑦

as a variable. To do so, note that the rectangle's corners lie on the ellipse, and so those

𝑥

and

𝑦

corner values are related by

4 𝑦 2 = 16 - 𝑥 2 𝑦 = \pm \sqrt 16 - 𝑥 2 4 = \pm \sqrt 16 - 𝑥 2 2

The positive square root corresponds to the top half of the ellipse (positive values of

𝑦

), while the negative square root corresponds to the lower half of the ellipse (negative values of

𝑦

). When we wrote the rectangle's height as

ℎ = 2 𝑦

we already assumed that

𝑦

is positive, so we'll just consider the positive square root in the following analysis.

We can now write the rectangle's area as

𝐴 = (2 𝑥) (2 \sqrt 16 - 𝑥 2 2) = 2 𝑥 \sqrt 16 - 𝑥 2

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.

𝑑 𝐴 𝑑 𝑥 = 𝑑 𝑑 𝑥 [2 𝑥 \sqrt 16 - 𝑥 2] = 2 [\sqrt 16 - 𝑥 2 + 𝑥 12 - 2 𝑥 \sqrt 16 - 𝑥 2] = 2 [16 - 𝑥 2 \sqrt 16 - 𝑥 2 - 𝑥 2 \sqrt 16 - 𝑥 2] = 2 16 - 2 𝑥 2 \sqrt 16 - 𝑥 2

5. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
The critical points occur when (I)

𝑑 𝐴 𝑑 𝑥

is undefined, or (II)

𝑑 𝐴 𝑑 𝑥 = 0

.

(I)

𝑑 𝐴 𝑑 𝑥

is undefined when the denominator is zero, or when

16 - 𝑥 2 = 0

, and so when

𝑥 = \pm 4

. Note, however, that those are the endpoints of the ellipse, and so

𝑦 = 0

there and the rectangle has no area. This critical point thus gives a minimum rather than a maximum area.

(II)When

𝑑 𝐴 𝑑 𝑥 = 0

16 - 2 𝑥 2 = 0 𝑥 2 = 8 𝑥 = \pm \sqrt 8 = \pm 2 \sqrt 2

6. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.
Since

𝑑 𝐴 𝑑 𝑥 = 2 16 - 2 𝑥 2 \sqrt 16 - 𝑥 2

we see that at our positive critical point

𝑥 = 2 \sqrt 2

(or, equivalently,

𝑥 2 = 8

𝑑 𝐴 𝑑 𝑥

goes from positive to negative, and hence

𝐴

is a maximum there.

Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .
Careful! The question asked for the maximum area that can be inscribed, and so to finish the problem we must provide that.

𝐴 m a x = 2 𝑥 \sqrt 16 - 𝑥 2 w i t h 𝑥 = \sqrt 8 = 2 \sqrt 8 \sqrt 16 - 8 = 2 \sqrt 8 \sqrt 8 = 2 \cdot 8 = 16 ✓

[close solution]

Rectangle with the greatest area is a square

Practice Problem: Rectangle with the greatest area is a square

Prove that of all rectangles with given perimeter

𝑃

, the square has the largest area.

1. Draw a picture of the physical situation.
Many students don't know where to begin when faced with this question. As always, we start with a figure: we're trying to demonstrate something about a rectangle, so we draw a rectangle, with sides labeled

𝑤

and

𝑙

for the width and length. See the figure.

The rectangle has a given perimeter

𝑃

, so

2 𝑤 + 2 𝑙 = 𝑃

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
We want to maximize the area of the rectangle,

𝐴 = 𝑙 𝑤

3. If necessary, use other given information to rewrite your equation in terms of a single variable.
We choose to write the area solely in terms of the width

𝑤

, and so we need to eliminate the length

𝑙

as a variable. To do so, note that we can rewrite the length in terms of the width and the given perimeter

𝑃

2 𝑤 + 2 𝑙 = 𝑃 𝑙 = 12 𝑃 - 𝑤

Thus we can write the area as

𝐴 = (12 𝑃 - 𝑤) 𝑤 = 12 𝑃 𝑤 - 𝑤 2

3. Take the derivative of your equation with respect to your single variable. Then find the critical points.
Remember that the perimeter

𝑃

is a given (it could be, say, 10 cm, or 23 inches; whatever), so it's a constant. For that given perimeter, we want to choose

𝑤

to maximize the rectangle's area, so we take the derivative with respect to

𝑤

𝑑 𝐴 𝑑 𝑤 = 𝑑 𝑑 𝑤 [12 𝑃 𝑤 - 𝑤 2] = 12 𝑃 - 2 𝑤

The critical points occur when

𝑑 𝐴 𝑑 𝑤 = 0

𝑑 𝐴 𝑑 𝑤 = 0 = 12 𝑃 - 2 𝑤 2 𝑤 = 12 𝑃 𝑤 = 14 𝑃

𝑑 𝐴 𝑑 𝑤 = 12 𝑃 - 2 𝑤

, as we increase the width from zero,

𝑑 𝐴 𝑑 𝑤

goes from positive to negative at

𝑤 = 14 𝑃

, and so by the first derivative test this value of

𝑤

corresponds to the maximum area for the rectangle.

When

𝑤 = 14 𝑃

𝑙 = 12 𝑃 - 𝑤 = 12 𝑃 - 14 𝑃 = 14 𝑃

Hence the maximum area of the rectangle occurs when the rectangle's width and length are equal to each other, and to one-quarter the perimeter of the total perimeter:

𝑤 = 𝑙 = 14 𝑃

. That is, the area is maximized when the rectangle has equal sides, and so is a square.

✓

[close solution]

Shortest ladder required to reach a house over a wall

Practice Problem: Shortest ladder required to reach a house over a wall

$As described in text$ A wall 10 feet high is six feet from a house. Find the length of the shortest ladder that will reach the house while leaning against the fence.
[ Hint: The math works out more easily if you call

𝑥

the distance from the base of the ladder to the wall (rather than to the house); see the figure. Even so, the algebra at the end can get messy. Don't spend more than a few minutes on that if you can't easily find the critical point(s); instead just check your work against ours up to that point, and then see a convenient way to factor.]

Answer: 22.39 feet

$As described in text$ 1. Draw a picture of the physical situation.
See the figure. As suggested by the hint, we've called the distance from the ladder's base to the wall

𝑥

, so the horizontal distance from the ladder's base to the house is

𝑥 + 6

. We're calling the ladder's height

𝑦

.

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
The ladder's length

𝐿

is given by

𝐿 2 = (𝑥 + 6) 2 + 𝑦 2

3. If necessary, use other given information to rewrite your equation in terms of a single variable.
We choose to write the ladder's length solely in terms of the variable

𝑥

, and so we need to eliminate

𝑦

. To do so, note that by similar triangles:

𝑦 10 = 𝑥 + 6 𝑥 𝑦 = (10) 𝑥 + 6 𝑥 = (10) (1 + 6 𝑥)

We can then rewrite our expression for

𝐿

solely in terms of

𝑥

𝐿 2 = (𝑥 + 6) 2 + (10) 2 (1 + 6 𝑥) 2

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.
The minimum value of

𝐿

occurs for the same value of

𝑥

as when

𝐿 2

is minimized, so let's define the function

𝑓 (𝑥) = 𝐿 2 = (𝑥 + 6) 2 + 100 (1 + 6 𝑥) 2

and minimize

𝑓

. (Otherwise we would write

𝐿

as the square root of everything on the right-hand side of the equation, but that just introduces a messy square root we then have to work with.)

𝑑 𝑓 𝑑 𝑥 = 𝑑 𝑑 𝑥 [(𝑥 + 6) 2 + 100 (1 + 6 𝑥) 2] = 2 (𝑥 + 6) + 2 (100) (1 + 6 𝑥) (- 6 𝑥 2) = 2 (𝑥 + 6) - 2 (600) (𝑥 + 6 𝑥) (1 𝑥 2) = 2 (𝑥 + 6) [1 - (600 𝑥 3)]

The minima occur when

𝑑 𝑓 𝑑 𝑥 = 0

𝑑 𝑓 𝑑 𝑥 = 0 = 2 (𝑥 + 6) [1 - (600 𝑥 3)]

𝑥 = - 6

is physically impossible. Hence our single critical point is

𝑥 = 3 \sqrt 600 = 8.43

ft.

5. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
6. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.

𝑑 𝑓 𝑑 𝑥

is negative for

𝑥 < 3 \sqrt 600

(try, say,

𝑥 = 1

), and is positive for

𝑥 > 3 \sqrt 600

(try, say,

𝑥 = 10

). Hence by the first derivative test

𝑥 = 3 \sqrt 600

gives us a minimum for

𝑓

, and hence for

𝐿 2

.

7. Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .
Careful! The question asked for the ladder's shortest length, so to finish we must compute that.

When

𝑥 = 3 \sqrt 600 = 8.43

ft,

𝐿 2 = (𝑥 + 6) 2 + 100 (1 + 6 𝑥) 2 = (8.43 + 6) 2 + 100 (1 + 6 8.43) 2 = 501.23 𝐿 = \sqrt 501.23 = 22.39 f t ✓

[close solution]

Rectangular & square enclosures

Practice Problem: Rectangular & square enclosures (based on an exam question)

A man has 340 yards of fencing to enclose two separate fields. One field will be a rectangle twice as long as it is wide, and must contain at least 800 square yards. The other field is a square and must contain at least 100 square yards.

(a)

𝑥

is the width of the rectangular field, what are the maximum and minimum possible values of x?

(b)

What is the greatest number of square yards that can be enclosed by the two fields? Justify your answer.

(a) $20 y a r d s \leq 𝑥 \leq 50 y a r d s$
(b) 5100 square yards

1. Draw a picture of the physical situation. $As described in text$ Also note any physical restrictions determined by the physical situation.
We found in part (a) that

20 y a r d s \leq 𝑥 \leq 50 y a r d s

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
The area of the two fields is given by

𝐴 t o t a l = 2 𝑥 2 + 𝑦 2

3. If necessary, use other given information to rewrite your equation in terms of a single variable.
Recall from part (a) that

𝑦 = 85 - 32 𝑥

, so we can write the total area just in terms of

𝑥

𝐴 t o t a l = 2 𝑥 2 + (85 - 32 𝑥) 2

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.

𝑑 𝐴 𝑑 𝑥 = 4 𝑥 + 2 (85 - 32 𝑥) (- 32) = 4 𝑥 - 3 (85 - 32 𝑥) = 4 𝑥 - 255 + 92 𝑥 = 172 𝑥 - 255

The critical value(s) of

𝑥

occur when

𝑑 𝐴 𝑑 𝑥 = 0

𝑑 𝐴 𝑑 𝑥 = 0 = 172 𝑥 - 255 𝑥 = 217 (255) = 30 y a r d s

5. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
To determine the maximum area, we need to compute the area for the critical value of

𝑥

we just found,

𝑥 = 30

yards, as well as at the endpoints we found in part (a),

𝑥 = 20

yards and

𝑥 = 50

yards.

First, for convenience, let's compute

𝑦

for each of those values of

𝑥

𝑦 = 85 - 32 𝑥 F o r 𝑥 = 20 : 𝑦 = 85 - 32 (20) = 55 y a r d s F o r 𝑥 = 30 : 𝑦 = 85 - 32 (30) = 40 y a r d s F o r 𝑥 = 50 : 𝑦 = 85 - 32 (50) = 10 y a r d s

Then computing the associated areas:

𝐴 t o t a l = 2 𝑥 2 + 𝑦 2 F o r 𝑥 = 20 : 𝐴 t o t a l = 2 (20) 2 + (55) 2 = 2 (400) + 3025 = 3825 s q u a r e y a r d s F o r 𝑥 = 30 : 𝐴 t o t a l = 2 (30) 2 + (40) 2 = 2 (900) + 1600 = 3400 s q u a r e y a r d s F o r 𝑥 = 50 : 𝐴 t o t a l = 2 (50) 2 + (10) 2 = 2 (2500) + 100 = 5100 s q u a r e y a r d s

Hence the greatest number of yards that can be enclosed occurs when

𝑥 = 50

yards, and is 5100 square yards.

✓

7. Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .
The question asked for the greatest number of square yards that can be enclosed, which we have provided.

✓

[close solution]

Triangle bounded by axes and curve

Practice Problem: Triangle in curve

A right triangle has one vertex at the origin and one vertex on the curve

𝑦 = 𝑒 - 𝑥 / 2

for

1 \leq 𝑥 \leq 4

. One of the two perpendicular sides is along the

𝑥

-axis; the other is parallel to the

𝑦

-axis. Find the maximum and minimum areas for such a triangle.
Note: You may use a calculator to compute the areas.

Answer: Minimum: 0.27; maximum: 0.37

$As described in text$ 1. Draw a picture of the physical situation.
See the upper figure. We had trouble envisioning this at first, but drawing the picture while working through the description makes it clear: One vertex is on the origin, so mark that spot first. One of the perpendicular sides is along the

𝑥

-axis, so draw that next. The other perpendicular side is parallel to the

𝑦

-axis, and must end on the curve, so draw that. We're showing it ending on the point

(𝑥, 𝑦)

on the curve.

Finally, the third side, the hypotenuse, must then go from that point back to the origin. See the lower figure.

2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.
The triangle has area

𝐴 = 12 (𝑏 𝑎 𝑠 𝑒) (ℎ 𝑒 𝑖 𝑔 ℎ 𝑡)

𝐴 = 12 𝑥 𝑦

3. If necessary, use other given information to rewrite your equation in terms of a single variable.
We choose to write the area solely in terms of the variable

𝑥

; hence we must eliminate

𝑦

. To do so, note that since the point

(𝑥, 𝑦)

lies on the curve,

𝑦

is related to

𝑥

𝑦 = 𝑒 - 𝑥 / 2

. Hence

𝐴 = 12 𝑥 𝑒 - 𝑥 / 2

4. Take the derivative of your equation with respect to your single variable. Then find the critical points.

𝑑 𝐴 𝑑 𝑥 = 𝑑 𝑑 𝑥 [12 𝑥 𝑒 - 𝑥 / 2] = 12 [𝑒 - 𝑥 / 2 + 𝑥 (- 12) 𝑒 - 𝑥 / 2] = 𝑒 - 𝑥 / 2 2 [1 - 12 𝑥]

The only critical point, when

𝑑 𝐴 𝑑 𝑡 = 0

, occurs when

𝑥 = 2

.

5. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.
We now just need to check the critical point (

𝑥 = 2

), and the endpoints (

𝑥 = 1

and

𝑥 = 4

), to find the minimum and maximum areas:

𝐴 = 12 𝑥 𝑒 - 𝑥 / 2 F o r 𝑥 = 1 : 𝐴 = 12 (1) 𝑒 - 1 / 2 \approx 0.30 F o r 𝑥 = 2 : 𝐴 = 12 (2) 𝑒 - 1 \approx 0.37 F o r 𝑥 = 4 : 𝐴 = 12 (4) 𝑒 - 2 \approx 0.27

Hence the minimum area is 0.27 (when

𝑥 = 4

), and the maximum area is 0.37 (when

𝑥 = 2

✓

7. Finally, check to make sure you have answered the question as asked: $𝑥$ or $𝑦$ values, or coordinates, or a maximum area, or a shortest time, or . . . .
The question asks for the minimum and maximum areas, which we have provided.

✓

[close solution]

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