3.7 Optimization Problems

After this lesson, you should be able to:

Be able to model the real-life problem to mathematical problem.Solve applied maximum and minimum problems.

One of the most valuable aspects of differential calculus is the ability to find where something is maximized or minimized. When you think about your own personal finances, aren't two of the most important questions you can ask "When/where do I have the least amount of cash and when/where do I have the most?" What factors led to that? At what point is this occurring and why?

Optimization Problems

Manufacturers want to maximize their profit while minimizing their costs. Thanks to calculus, we've learned that our derivative tells us plenty about our function. And when our derivative is zero, WE WANT TO PAY CLOSE ATTENTION TO WHAT IS HAPPENING TO OUR FUNCTION. Remember that max's and min's can occur where our derivative is zero or undefined, so applying this concept is not too difficult. Now, let's define the steps in analyzing optimization problems.

Optimization Problems

Example

Example 1 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?

hxhxV 2),(

Solution

Let the length of the base be x inch,

xx

hand the height be h inch.

Then the volume of the box in terms of x and h inch is

Primary EquationNotice that “surface area of 108

in2 ” provides 1084),( 2 xhxhxS Secondary Equation

Example

Example 1 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume?

xxh 4/)108( 2

Solution

From the secondary equation, we have

xx

h

Then the volume of the box in terms of x in the primary equation is

Now, before we aim on V(x) and try to find its maximum volume while x has certain length, we should find the feasible domain of x.

4/274/)108()( 322 xxxxxxV

Notice that 22 4108),( xxhxhxS

This means 1080 x

0xand

Then taking the derivative of V(x) and set to zero. We have 04/327)(' 2 xxV

6xThe critical number within the feasible domain is

Feasible Domain

Compute the endpoints 0)0( V

and

108)6( V

0)108( Vand Therefore, V(6, 3) is maximum when x = 6

and h = 3. The box dimension is 6 x 6 x 3.

324/)36108(4/)108( 2 xxh

Guidelines for Solving Applied Minimum and Maximum

Problems

Example

Example 2 You are designing an open box to be made of cardboard that is 10 inches by 15 inches. The box will be formed by making the cuts at each corner. How long should you make the cuts? What is the maximum volume?

15

10

Example

x215

Solution

Let the side of square to be cut be x inch.

Then the dimensions of the boxd can be interpreted asLength =Therefore, the volume is

xxxxV )210)(215()(

15

10

xx

xx

xx

xx

xxx 150504 23

x210 Width = xHeight =

Notice that 102 x

Feasible Domain

0x and

50 xThen

Then taking the 1st derivative of V(x) and set to zero. We have

015010012)(' 2 xxxVThe critical number within the feasible domain is

xxxxV 150504)( 23

Example

6

7525 x

domain)

feasible thein not is 56

7525(

x

27

)7710(125

6

7525

V

Then taking the 2nd derivative of V(x), we have 10024)('' xxV

Therefore, is maximum

volume when we cut 4 squares of side .

15010012)(' 2 xxxV

Example

07201006

752524

6

7525''

V

27

)7710(125

6

7525

V

6

7525

Example

Example 3 Which points on the graph of y = 4 – x2 are closest to the point (0, 2)?

x

y

d (x, y)

Solution

The distance between the point (0, 2) to any point (x, y) on the graph is 2222 )2()2()0( yxyxd

Notice that y = 4 – x2, then

43)24()( 24222 xxxxxd

Let D(x) = d2(x), then43)( 24 xxxD

Then taking the 1st derivative of D(x) and set to zero. We have

0)32(264)(' 23 xxxxxDThe critical number within the feasible domain is

Example

0x and 2

6x

Then taking the 2nd derivative of D(x), we have

060'' D

)12(6612)('' 22 xxxD

0212

32

2

6''

Dan

d This verifies that x = 0 yields a relative maximum andyield a minimum distance to the given point.

2

6x

x

y

Notice that in this example, we can not find the absolute maximum, but only the relative maximum. However, we do have absolute minimum.

43)( 24 xxxD

Example

Example 4 A man is in a boat 2 miles from the nearest point R on the coast. He is trying to go to a point Q, located 3 miles down the coast and 1 mile inland. He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time?

Q

2

1

3

Solution

Let the point on the coast toward by the man be P, and PR = x, then PS = 3 – x. Then

PR Sx

3 – x

M

42 xMP

1)3( 2 xPQ

Example

Q

2

1

3Then the total time from point M to Q will be

PR Sx

3 – x

M

4

106

2

4)(

22

xxxxT

01064

3

42 22

xx

x

x

x

dx

dT

Then taking the 1st derivative of T(x) and set to zero. We have

)106(4

96

4 2

2

2

2

xx

xx

x

x012896 234 xxxx

The feasible domain for x is [0, 3]. The solution for x within the domain is x = 1.

x

y

Then compute

Example

4

101)0( T

4

53)1( T

4

1321)3(

T

We conclude that when x = 1 yields the minimum time.

Example

Example 5 Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum area?Solution

r

x

x

Let the side of the square be x and the radius of the circle be r. Then

Since the total length of wire is 4 feet, then

22),( rxrxA Primary Equation

Secondary Equation

424 rx So,

)1(2 xr

Example

r

x

x

The feasible domain of x is restricted by the square’s perimeter

22 )1(2

)(

xxxA

)1(2 x

r

22 )1(4 xx

]48)4[(1 2 xx

10 x

08)4(2

x

dx

dATaking the 1st derivative of A(x) and set to zero. We have

The critical number within the feasible domain is

4

4

x

Then compute

4

)0( A

4

4)4

4(

A

1)1( A

We conclude that when yields the maximum area. That means all the wire is used to form circle.

Example

0x

x

y

Example

Example 6 A square fountain has a water pool of d feet wide around. A repairman tried to fix a water pipe in the fountain. He used 2 pieces of wood of length L(<d) to access the water pipe in the center. What is the minimum length of the wood board? How did he place these 2 pieces of wood?SolutionWe simplify the problem to the following question: in a square of side length of d, find the 2 perpendicular segments of length L and when L can reach its minimum.

d

dLL

Example

Let the acute angle between one wood board and outer edge of the fountain be θ. Extending TS to R.

d

dL

L

Q

S

T

R P

Then in TRQ, RQ = d tan And PQ = d – Lcos

Therefore RP = RQ – PQ = d tan – (d – Lcos )

= d (tan – 1) + Lcos In PRS, RS = RP sin =[d (tan – 1) + Lcos ]sin In TRQ, RS + ST = d/cos Or, [d (tan – 1) + Lcos ]sin + L = d/cos Solve for L, we get

cossin1

cossin)(

dL

ExampleNow we seek what value will yields the minimum L. Or,We only need what value will yields the minimum value to f( ). We notice that 0 < < /2 is the feasible domain.

)(cossin1

cossin)(

dfdL

Taking the 1st derivative of f( ), we have

2

2

)cossin1(

])cos(sincossin1)[sin(cos

2)cossin1(

2cos)cos(sin)cossin1)(sin(cos)('

f

2)cossin1(

)cossin)(sin(cos

2)cossin1(

)cos)(sin2)(sin2/1(

The critical number within the feasible domain is

4

0( 2

and are not in the 0 < < /2 )

ExampleWe are going to take the 2nd derivative of f( ). The derivative of the numerator is

4

2

)cossin1(

)cossin1)(1cossin3)(cos(sin)(''

f

2)cossin1(

)cos)(sin2)(sin2/1()('

f

d

d )]cos(sin2sin2/1[

)cos(sincossin)cos(sin2cos ]cossin)cos(sin)[cos(sin 2

)1cossin3)(cos(sin

4)cossin1(

2cos)cossin1(2)cos)(sin2)(sin2/1(

ExampleWe simplify the 2nd derivative of f( ). We get

4

2

)cossin1(

)cossin1)(1cossin3)(cos(sin)(''

f

2)cossin1(

)cos)(sin2)(sin2/1()('

f

4)cossin1(

2cos)cossin1(2)cos)(sin2)(sin2/1(

4

2

)cossin1(

)cossin1)(1cossin3)(cos(sin

4)cossin1(

)cossin1)(cos(sin2cos2sin

4

2

)cossin1(

)cossin1()sin)(coscos(sin2sin

Example

4)cossin1(

)cossin1)(cos(sin)(''

f

])sin(cos2sin)cossin1)(1cossin3[( 2

4)cossin1(

)cossin1)(cos(sin

)2sin1(2sin2sin

2

1112sin

2

3

When =/4, we know that all the factors are positive. So 0)

4(''

f

When =/4, f( ) yields the minimum value. Therefore, ddfL

3

22)4()(

)(cossin1

cossin)(

dfdL

x

y

Homework

Pg. 223 3-9 odd, 17 – 23 odd, 27, 29, 33, 49