Lesson 17 - Optimization Problems, Slides of Calculus for Engineers

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Lesson 6

Optimization Problems

(Maxima and Minima

Problems)

OBJECTIVES:

• to identify the quantity to be maximized or

minimized,

• to apply the knowledge of derivatives and

critical points in solving maximum and minimum

problems and

• to solve maximum and minimum problems with

ease and accuracy.

Example 1.

Divide 120 into two parts such that the product of one part and

the square of the other is a maximum. Find the numbers.

 

    

2

2 2

2

2

dM 240x- 3 x

dM 240x- 2 x x

dM 120 - x 2 x x 1

Equation: M 120 - x x

M bethe productof one partandthesquareof theother part

120 - x betheother part

Let x beone part

 



80 x 0 x 80

3 x 0 x 0 3x 80 - x 0

0 240x-3x 0 dx

dM (^2)

   

   

answer: 80 and 120 - x 40

when 0 ; 240 6 0 minimum

when 80 ; 240 6 80 maximum

Using the 2 ndderivativetest 240 6

 

     

     

 

x

dx

d M

x

dx

d M

x

x

dx

d M

16 - 2x

x

   

  

2

2 3

2

160 - 104x 12x

dx

dV

160x-52x 4 x

h height x 160 - 52x 4 x x

W width 10 - 2x 16 - 2x 10 2 x x

Let L length 16 - 2x V LWh

  

thus x 2

6. 67

3

23 x 2 and x

x- 2 3 x 20 0

3x -26x 40 0

0 160 - 104x-12x 0

dx

dV

2

2



  

 

 

  

x must not be equal to 6. 67 because

10 - 2 x will be negative

W widthoftherectangle 2 y

Let L length oftherectangle 2x

 

 

2 2

2

2 2

4 100 x 100 x

4 x A'

100 x 4 2 100 x

2 x A' 4 x

 

x 50 5 2 in.

x 50

8 x 400

4 x 400 4 x 0

0 100 x

4 x 4100 x

4 100 x 0 100 x

4x A' 0

2

2

2 2

2

2 2

2 2

2

 



 

   

 

  

   

  

   

  

2

2

A 200 in.

therefore, A 4 50 100 50

2

2 2

2 2

2 2

2 2 2

2 2 2

A 4x 100 - x

we have, A 4x R x

substituting y R x to A 4xy

y R x

then y R x

since,R x y

A 4xy

A LW 2x 2 y

Example 4.

Find the altitude of the largest circular cylinder that can

be inscribed in a circular cone of radius R= 5 ” and

height H= 10 ”.

H=

”

10 -

y

y

x

R=5”

Example 5.

A rectangular field of fixed area is to be enclosed and divided into

three lots by parallels to one of the sides. What would be the

relative dimensions of the field to make the amount of fencing a

minimum?

L

W

A area of the field

Let P the amount of fence needed to enclose the lots





L

2 L 4 A

P

L

A

then, P 2L 4

substituting W in P 2 L 4 W;

L

A

P 2 L 4 W and A LW W

2

    

2

2

2

2 2

2

2

L

2 L 4 A

L

4 L 2 L 4 A

dL

dP

L

L 4 L 2 L 4 A 1

dL

dP

but A is constant,hence

L

therefore, L 2 W or W

2 L 4 LW

2 L 4 A but, A LW

L

2 L 4 A

dL

dP

2

2

2

2

 

2

2

r

V

then, h

V rh Volume of cylinder

r

V

T 2 r 2

r

V

2 r 2 r

T 2 r 2 rh

Solving for the Total Surface Area T :

2

2

2

2

2

2

r

2 V

4 r dr

dT

r

V 1

4 r 2 dr

dT

d h 2

h

2

d

2

d since, r 2

h r

2

rh r

but, V rh 2

V

4

2 V r

r

2 V 4 r

0 r

2 V 0 4 r dr

dT

2 3

3 2

2

2

  

 



 

 

 



 

    

Example 7.

Find the proportion of the circular cylinder of largest

volume that can be inscribed in a given sphere.

r

a

h 2a

h/

C

a

r