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§12.5 Absolute Maxima and Minima.

12.6 Optimization

The student will learn about:

absolute maxima,

absolute minima, and

the second derivative and extrema.

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Absolute Maxima and Minima.

Definition: f (c) is an absolute maxima of f if f (c) ≥ f (x) for all x in the domain of f.

Definition: f (c) is an absolute minima of f if f (c) ≤ f (x) for all x in the domain of f.

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Example 1

Graphing calculator

Window 0 x 20

0 y 40.

Using the graph utility “minimum” to get -

x = 3 and y = 18.

x

27x3)x(f

Find the absolute minimum value of,

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Extreme Value Theorem

Theorem 1. Extreme Value Theorem.

A function f that is continuous on a closed interval [a, b] has both an absolute maximum value and an absolute minimum value on that interval.

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Steps in finding absolute maximum and minimum values

Theorem 2. Absolute extrema (if they exist) must always occur at critical values of the derivative or at end points.

a. Check to make sure f is continuous over [a, b].

b. Find the critical values in the interval [a, b].

c. Evaluate f at the end points a and b and at the critical values found in step b.

d. The absolute maximum of f (x) on [a, b] is the largest of the values found in step c.

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Example 2

a. The function is continuous.

b. f ‘ (x) = 3x 2 – 12x = 3x (x – 4). Thus critical values of 0 and 4.c. f (-1) = - 7, f (0) = 0, f (4) = - 32, and f (7) = 49

Find the absolute maximum and absolute minimum value on [-1, 7] of,

23 x6x)x(f

d. The absolute maximum is 49.

e. The absolute minimum is - 32.

The above may be confirmed with a graphing calculator

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Second Derivative Test

f ‘ (c) f “ (c) Graph of f is f (c)

0 +

f ‘ (c) f “ (c) Graph of f is f (c)

0 + Concave up Local minimum

0 -

f ‘ (c) f “ (c) Graph of f is f (c)

0 + Concave up Local minimum

0 - Concave down Local maximum

0 0

f ‘ (c) f “ (c) Graph of f is f (c)

0 + Concave up Local minimum

0 - Concave down Local maximum

0 0 ? Test fails

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Example 2 continued

b. Thus critical values of 0 and 4.

f “ (0) =

Find the local maximum and minimum value of,

23 x6x)x(f

The above may be confirmed with a graphing calculator

a. f ‘ (x) =

b. Thus critical values of 0 and 4.

f “ (0) = -12 hence f (0) local maxima.

f “ (4) =

b. Thus critical values of 0 and 4.

f “ (0) = -12 hence f (0) local maxima.

f “ (4) = 12 hence f (4) local minimum.

a. f ‘ (x) = 3x 2 – 12x = 3x (x – 4).

f “ (x) =

a. f ‘ (x) = 3x 2 – 12x = 3x (x – 4).

f “ (x) = 6x – 12 = 6 (x – 2)

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Second Derivative Test

Theorem 3. Let f be continuous on interval I with only one critical value c in I.

If f ‘ (c) = 0 and f “ (c) > 0, then f (c) is the absolute minimum of f on I.

If f ‘ (c) = 0 and f “ (c) < 0, then f (c) is the absolute maximum of f on I.

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Summary.

• All continuous functions have absolute maximum and minimum values.

• These absolute values will be found at either critical values or end points of the intervals on which the function is defined.

• Local maxima and minima may also be found using these methods.

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§12.6 Optimization

The student will learn about:

area and perimeter,

revenue and profit, and

inventory control.

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Area and Perimeter – Example 1

The techniques used to solve optimization problems are best illustrated through examples. Let’s begin with a geometric example.

Find the dimensions of a rectangular area of 225 square meters that has the least perimeter.

Drawing a figure might help.

l

w

A = l · w = 225

P = 2 · l + 2 · w

From the area equation solve for l and substitute that value of l into the perimeter equation to get an equation in two unknown.

w

225l

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Example 1 - continued

Find the dimensions of a rectangular area of 225 square meters that has the least perimeter.

We wish to minimize P so we will take the derivative and look at the critical values.

P = 2 · l + 2 · w

There is a critical value at 15. (Disregard – 15 since the distance cannot be negative.)

w

225l

P = 2 · l + 2 · w = w2w

450

22

2

w

)15x()15x(2

w

450w2)w('P

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Example 1 - continued

Find the dimensions of a rectangular area of 225 square meters that has the least perimeter.

P ” (15) > 0 thus the least perimeter occurs when w = 15.

3w

900)w("P

When w = 15, l is also 15, hence the shape is a square of side 15 meters with minimum perimeter of 60.

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Optimization Strategies

1. Introduce variables, look for relationships among these variables, and construct a math model of the form

Maximize (minimize) f (x) on the interval I.

1. Introduce variables, look for relationships among these variables, and construct a math model of the form

Maximize (minimize) f (x) on the interval I.

2. Find the critical values of f (x).

1. Introduce variables, look for relationships among these variables, and construct a math model of the form

Maximize (minimize) f (x) on the interval I.

2. Find the critical values of f (x).

3. Use section 4-4 to find the maximum (minimum) value of f (x) on the interval I.

1. Introduce variables, look for relationships among these variables, and construct a math model of the form

Maximize (minimize) f (x) on the interval I.

2. Find the critical values of f (x).

3. Use section 4-4 to find the maximum (minimum) value of f (x) on the interval I.

4. Use the solution to the mathematical model to answer all the questions asked in the problem.

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Maximizing Revenue and ProfitA company manufactures and sells x television sets per month. The monthly cost and price-demand equations are:

R (x) = xp =

C (x) = 60,000 + 60x and p = 200 – x/50 for 0 ≤ x ≤ 6,000

a. Find the maximum revenue.

Maximum at R ‘ (x) = 0

50200

2xx

R ‘ (x) = 25

200x

or x = 5,000, and

R(5000) = $500,000

0 x 6,000 0 y 600,000.

R (x)

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Application #1 continued

P (x) =

b. Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set.

P ‘ (x) has a critical value at x =

000,6014050

2

xx

P (3500) =

p (3500) =

C (x) = 60,000 + 60x and p = 200 – x/50 for 0 ≤ x ≤ 6,000

P ‘ (x) = 14025

x

3500.

$185,000

$130 per TV.0 x 6,000 0 y 200,000.

P (x)

50200

2xx R (x) =

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Application #2

A 300 room hotel in Las Vegas is filled to capacity every night at $80 per room. For each additional $1 increase in rent, 3 fewer rooms are rented. If each rented room cost $10 to service per day, how much should the management charge for each room to maximize gross profit. What is the maximum gross profit?

Sometimes it is difficult to come up with the appropriate equations in order to solve a problem. For some of you that might be true in this case. I will show you a method of organizing your data to assist you in coming up with the equations needed.

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Application #2 continued

300 room filled to capacity at $80 per room. For a $1 increase in rent, 3 fewer rooms rent. Each room cost $10 to service per day. How much to charge for each room to maximize gross profit. What is the maximum gross profit?

Solution by Table. Let x be the amount over $80 for a room.x $/room # rooms Revenue Cost Profit0 80 300 1 2 3 4

x

R (x) = - 3x2 + 60x +24000

C (x) = 3000 – 30x and P (x) = - 3x2 + 90x + 21,000

81 82 83 84

297 294 291 288

x + 80

300 – 3x

$24,000 $24,057 $24,108 $24,153 $24,192

$3,000 $2,970 $2,940 $2,910 $2,880

$21,000 $21,087 $21,168 $21,243 $21,312

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Application #2 continued

300 room filled to capacity at $80 per room. For a $1 increase in rent, 3 fewer rooms rent. Each room cost $10 to service per day. How much to charge for each room to maximize gross profit. What is the maximum gross profit?

P (x) = The maximum profit occurs when P ‘ (x) is at a critical value.

P ‘ (x) = - 6x + 90 so

x = 15, so the room

- 3x2 + 90x + 21,000

charge should be $95, and

P (15) = $21,6750 x 25 21,000 y 22,000.

P (x)

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Inventory Control

A pharmacy has a uniform annual demand for 200 bottles of a certain antibiotic. It cost $10 to store one bottle for one year and $40 to place an order. How many times during the year should the pharmacy order the antibiotic in order to minimize total cost.

To calculate the total cost we use 40y for the order cost and 5x for the storage cost. We use 5 instead of 10 because the storage cost average between 0 and 10 based on the storage cost of a bottle that is sold when it first arrives and the cost of a bottle stored for the full time before the next order.

Let x = number of bottles purchased with each order, and y = number of orders.

Hence, C = 40y + 5x.

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Inventory Control continued

Minimize total cost of 200 bottles where x = number of bottles per order, and y = number of orders. C (x) = 40y + 5x.

To write C in terms of one variable we need a relationship between x and y. But the total number of bottles ordered is 200 and that is xy. So y = 200/x.

And, C (x) = 8000/x + 5x.

C ‘ (x) = -8000/x 2 + 5

Set C’ (x) = 0 and we get the equation 5x 2 = 8000

x 2 =x 2 = 1600 and x =x 2 = 1600 and x = 40Cost are minimized with five orders of 40 bottles.

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Summary.

• We saw how to optimize geometric problems.

• We saw how to optimize revenue and profit problems.

• We saw how to optimize inventory control problems.