The goal of this set of supplemental notes is to provide you with some more worked examples of optimization problems.

Remember the steps in problem solving:

1.

Read the problem. When doing this, determine what is given and what needs to be found.

2.

Develop and equation. (Sometimes a picture will help you do this.)

3.

Solve

4.

Check to see if the answer makes sense.

EXAMPLE 1:

Find two positive real numbers x and y such that their sum is 50 and their product is as large as possible.

SOLUTION:

Two positive real numbers x and y Þ answers will positive numbers.

Their sum is 50 Þ x + y = 50 or y = 50 - x.

Product as large as possible Þ find the maximum.

Product Þ xy.

P = xy = x (50 - x) = 50x - x ²

Find the first and second derivatives.

P' = 50 - 2x

P'' = -2

Set P' = 0 to find the critical point.

P' = 50 - 2x = 0 ® 2x = 50 ® x = 25

Since P'' < 0, which means the function is concave down, by the second derivative test, x = 25 is a maximum. If x = 25, then y = 50 - x = 25, and the product is 625.

Does the answer make sense?

Yes, the answers are positive numbers that sum up to 50.

EXAMPLE 2:

A farmer has 600 m of fencing with which he plans to enclose a rectangular pen adjacent to a long an existing wall. He will use the wall for one side of the pen and the available fencing for the remaining three sides. What is the maximum area he can enclose in this way?

SOLUTION:

Total of 600 m of fencing.

Going to use an existing wall as one of the sides.

Use the fencing for the other 3.

Maximize the area.

Develop the equation.

A = x (600 - 2x) = 600x - 2x ²

Find the first and second derivatives.

A' = 600 - 4x

A'' = -4

 Set A' = 0 to find the critical point.

A' = 600 - 4x = 0 ® 4x = 600 ® x = 150 m.

Since A'' < 0, the second derivative test tells us that x is a maximum. The maximum area is 45000 m².

The area is positive and so are the lengths of the sides, so it does make sense.

EXAMPLE 3:

A rectangular box has a square base with edge at least 1 in. long. It has no top, and the total area of its five sides is 300 in ². What is the maximum possible volume of such a box?

SOLUTION:

Height is at least 1 in. long.

No top.

Total area of the five sides is 300 in ².

Maximize the volume.

Using the fact that the total area of the five sides is 300 in ², we must first determine the equation for h.

Recall that the volume of a box is defined to be V = (length)(width)(height), so the volume of this box will be the following.

Find V' and V''.

Set V' = 0 and solve for x.

x cannot be a -10, so that is why I have not included it.

Since V'' < 0, the second derivative test tells us that x = 10 in a local maximum. If x = 10 in., then the height if 5 in., and the volume is 500 in ³.

Yes, all the dimensions are positive and the height is greater than 1 in.

EXAMPLE 4:

A commuter train carries 600 passengers each day from a suburb to a city. It costs $1.50 per person to ride the train. If is found that 40 fewer people will ride the train for each 5¢ increase in the fare. What fare should be charged to make the largest possible revenue?

SOLUTION:

Train carries 600 passengers.

It costs $1.50 per person to ride the train.

For every 5¢ increase in the fare, 40 fewer people will ride.

What fare should be charged to maximize the revenue.

First of all, revenue is equal to the price times the quantity. Therefore, Revenue = (price of the fare)(number of people riding the train). Let x be the number of 5-cent price increases. 150 - 5x will represent the new fare and 600 - 40x will take into account the decrease in the number of people riding the train after the price increase.

Find R' and R''.

R' = -3000 - 400x

R'' = -400

Set R' = 0, and solve for x. Remember that the price increase must be an integer.

-3000 - 400x = 0 ® -400x = 3000 ® x = -7.5

Since x = -7.5, then the two nearest integers are x = -7 and x = -8. Since R'' < 0, then x is a relative maximum. When x = -7, the fare will be $1.15 and the revenue will be $1012.00. When x = -8, the fare will be $1.10 and the revenue will be $1012.00.

Why didn't we use x = -7.5? Do we deal in fraction of a cent?

Yes, it does. First of all, we are not dealing with fraction of a cent, and if we want to maximize the revenue, the fare must go down.

EXAMPLE 5:

The sum of two positive numbers is 48. What is the smallest possible value of the sum of their squares?

Two positive numbers Þ The answer will be two positive numbers, say x and y.

The sum of two positive numbers is 48 Þ x + y = 48 or y = 48 - x

Smallest possible value Þ we are looking for a minimum

Sum of their squares Þ x ² + (48 - x) ²

S = x ² + (48 - x) ² = 2x ² - 96x + 2304

Find the first and second derivatives.

S' = 4x - 96

S'' = 4

Set S' = 0 to find the critical point.

S' = 4x - 96 = 0 ® 4x = 96 ® x = 24

Since S'' > 0, which means that the function is concave up, by the second derivative test, x = 24 is a minimum. If x = 24, then y = 48 - 24 = 24, and the sum of the squares is 1152.

Yes, x and y are positive and their sum is 48.

EXAMPLE 6:

A rectangle of fixed perimeter 36 is rotated about one of its sides to generate a right circular cylinder. What is the maximum possible volume of the cylinder?

P = 2x + 2y = 36 ® x + y = 18 ® y = 18 - x

Volume of a right circular cylinder is V = p r ² h.

V = p (18 - x) ² x = p (324 - 36x + x ²) x = p (324x - 36x ² + x ³)

Find the first and second derivatives.

V' = p (324 - 72x + 3x ² )

V'' = p (-72 + 6x)

Set V' = 0 to find the critical points.

p (324 - 72x + 3x ²) = 0 ® 108 - 24x + x ² = 0 ® (x - 18)(x - 6) = 0 ® x = 6 or x = 18

When x = 6, V'' (6) = p (6(6) - 72) < 0, therefore it is concave down.

When x = 18, V'' (18) = p (6(18) - 72) > 0, therefore it is concave up.

Since we want to maximize the volume, we want the second derivative to be concave down at the max value. This happens at x = 6, so y = 18 - 6 = 12, and the volume is 864p .

Yes, it does. If x = 18, then we would not have had a rectangle. So it had to be the other value.

EXAMPLE 7:

Show that among all rectangles with a given perimeter, the one with the largest area is a square.

All rectangles with a given perimeter, p

Find the first and second derivatives.

Set A' = 0 to find the critical point.

Since A'' < 0, which means the function is concave down, by the second derivative test, x = p/4 is a maximum.

So it is a square that has the largest area.