Lecture 11

Max-Min Problems

Maxima and Minima

Problems of type:

“find the largest, smallest, fastest, greatest, least, loudest, oldest, youngest, cheapest, most expensive, … etc.

If problem can be phrased as in terms of finding the largest value of a function then one is looking for the highest and lowest points on a graph(if they exist)

There is not a one-step test to detect “highest” and lowest points on a graph. What we can do is detect where relative

maxima and minima occur. Relative extrema occur atcritical numbers (rememberthat end points are criticalnumbers)

Strategy for solving Max-Min:• Phrase problem as a function for which one is to

find largest or smallest value• Find all critical numbers of the function (including

end points)• Make a table of values of the function at the critical

numbers• IF the function is defined on a closed interval then

the largest (smallest) functional value in the table is the maximum (minimum) value of the function.

• If not on a closed interval can still detect the local maxima and minima – may be able to determine if one of them is an absolute max or min by sketching the graph.

x

0

2

length = xlength = f(x) = 8 x

3

P(x) = 2 x + 2( )

8 x3

8 x3

P(x) 2 x 16 2 x3

P ' (x) 2 6 x2

0 x 2

Critical numbers = 0,2,1

33

x P(x)

0 16

2 4

1

33 4

93 16

Max Value

Function is on closed interval [0,2]

Want x-value of highest point

x ( )3 x 2f ' 3 x2

2 x =

Critical numbers = -1 (ep), 0 (ep), -2/3

x f(x)

0 0

-1 0

-2

3

4

27MAX

Function is defined on closed interval somax value at critical number is max value offunction.

x

Length = f(x) – g(x) or g(x)-f(x)

h(x) = f(x) – g(x)

( )h x x3

2 x2

3 2 x-1.5 x 1.5-1.5

h ' (x) 3 x2

4 x 2

0 3 x2

4 x 2

x 2

3

1

310 or x 2

3

1

310

x 1.720759220 (> 1.5) or x -.3874258863

Critical numbers = -1.5 (ep), 1.5 (ep), 2

3

10

3

x h (x)

-1.5 -1.875

1.5 -1.125

.38742 3.4165 MAX

Just because aderivative is 0 does notmean it is a critical numberof the function underconsideration – Here1.720359.. is not in the domain.

Length x

Length y

vol x2

y 10 x2

y y10

x2

Area = base + 4 sides = x*x + 4*x*y

Area = base + 4 sides = x*x + 4*x*y

( )A x x2 4 x 10

x2

( )A x x2 40

x

A ' (x) 2 x40

x2

A ' (x) 2x3

20

x2

Looking for lowest point –will occur at x = critical numbernear 3.

A ‘ = 0 if x = 20

1

3 = 2.7144

( )A 2.71441 22.1042This is an example wherethe domain is not a closedInterval but we can stilldetermine that the minoccurs at the one critical number

You can buy any amount of motor oil at $.50 per quart. At $1.10 you can sell 1000 quarts but for each penny increase in the selling price you will sell 25 fewer quarts. Your fixed costs are $100 regardless of how many quarts you sell. At what price should you sell oil in order to maximize your profit. What will be your maximum possible profit?

Profit = Income - Costs= (number sold)*(selling price) – [ (number purchased)*(purchase price) + fixed costs]

Let x = increase in price in pennies

Profit = (1000 -25*x)(1.10+.01*x) - [ (1000-25*x)*.50 + 100]

( )p x 500.00 985.00 x 25 x2

0 x 40

Critical numbers = 0 (ep), 40 (ep), -5/.5 = 10

Optimum selling price = $1.10 + (-.01*10) = $1

p ' (x) 5.00 .50 x

p ' (x) 500.00 5.00 x .25 x2

x p(x)

0 500.00

40 -100.0

10 525.00

Maximum profit

We determine herethat there is no needto consider x > 40 whichgives us a closed intervalto work on