4.7

Optimization Problems

Steps to follow for max and min problems.

(a) Draw a diagram, if possible.

(b) Assign symbols to unknown quantities.

(c) Assign a symbol,Q, to the quantity tobe maximized or minimized.

(d) Express Q in terms of the other symbols.

(e) Eliminate all but one unknown symbol,say x, from Q.

(f) Find the absolute maximum or minimumof Q = f (x).

i State the domain of .f x

ii Find the critical number of .f x

test to determine whether has af xmax or a min at the critical number.

value of occur at an end point?f x

(iii) Use the first derivative

(iv) Does the absolute max or min

1. Example: Find two numbers whose difference is 100 and whose productis a minimum.

Solution:Let denote one of the numbersxand the other.y

x y 100 or 100.y x Put P xy

=˃p(x)=x(x-100)

(i) Domain of P(x): all real numbers.

(ii) Critical numbers of P(x) 1 100 1x x

2 100x

0 x = 50.(iii) First Derivative Test

50

50( )P x

P′(x)=

P′(x)

(iv) When 50,x

50 100 y y 50.

x y50 50, .

P(50) is an absolute min value.

Graph: ( ) ( 100)P x x x

2. Example: A farmer wants to fence arectangular enclosure for his horsesand then divide it into thirds with fencesparallel to one side of the rectangle. If he has 2000m of fencing, find the areaof the largest rectangle that can be enclosed.Solution:Let be the length and thel w

width of the rectangle.

l

w

A lwP 4 2 2000w l

w 2000 2

4

l

A l 2000 2

4

ll

1000

2

ll

(i) Domain of A(l):

0 1000.l

2

5002

ll

(ii) Critical numbers of A(l)

A l 500 l

A l 0 l 500.

By the first derivative test we can show that A(500) is the max value of A,

2000 2

4

lw

2000 1000

4

250.

500 2502125,000 meters

When l=500,

max area of the rectangle =

Graph:2

A( ) 5002

ll l

3. Example:Find the point on the parabola

Let D denote the distance from (0,3) to

any point (x,y) on the parabola.

x=y2 that is closest to the poit (0,3).

D 2 2( 0) ( 3)x y

2 2 2( ) ( 3)y y

4 2( 3)y y

a minimum.

(i) Domain of D(y): all real numbers > 0.

D(y)=

Find the value of y which makes D(y)

(ii) Critical numbers:

3

4 2

2 3

( 3)

y y

y y

2

4 2

( 1)(2 2 3)

( 3)

y y y

y y

0 y 1, since 2 2 3 02y y

for all y.

D′(y)

(iii) First Derivative Test

1

has an absolute minimum when y 1.

When 1,y x y 2

1,1 is the required point

1.

2on the graph of .x y

D′(y)

By the first derivative test D(y)

Graph: 4 2( ) ( 3)D y y y

4. Example: A right circular cylinder isenscribed in a sphere of radius . Find therlargest possible volume of such a cylinder.

Solution:Let be the radius,x

be the height and2yV the volume of the

y

x

rcylinder.

V x y2 2( ).

2V 2x y

x y r2 2 2 2 2 2x r y

V( )y 2 2 2r y y

(i) Domain of V(y): 0 y r

V ( )y V ( ) 0 y 3 2 2y r

.3

ry

(ii) Critical numbers of V(y)

Note that V 0 V 0.r

2π r2 -3y2

By the first derivative test

That is V is a maximum where .3

ry y

Maximum value for V( ) isy3

223 3

r rr

3 3

23 3 3

r r

V3

r

3 332

3 3

r r

322

3 3

r34

.3 3

r

5. Example: Mountain Beer sells its beerin an aluminum can in the shape of a right-circular cylinder. The volume of each can

What should the dimension of the can be in order to minimize the amountof aluminum used?

Solution:Let be the height, the radius, theh r V

volume and the surface area of the can.S

V 2r h 250

h2

250

r

2

250

r

is ml.250

2 2 2 r h rS and

2 r

h

r

S r h 2 r2 r2

2 2 2 r h r

Surface Area,

Volume Area of base height2r h 2 .r h

22 r 2

2502 r

r 2 500

2 rr

(i) Domain of S(r) r 0

(ii) Critical numbers of S(r)

2

5004 r

r

3

2

4 500r

r

2 2 2 r h rS

S(r)=

S′(r)=

3 125r 5cm(iii) First Derivative Test 3

2

4 125r

r

0 5

0 5

when 5cm.r

S′(r)=0 =˃

S′(r)=

S′(r)

S(r)

S(r) has its absolute minimum value

when r 5,2

250h

r

250

25

10cm

The surface area is a minimumwhen 5cm and 10cm.r h

Graph: 2 500( ) 2S r r

r