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# OPTIMIZATION PROBLEMS

## (MAXIMA and MINIMA PROBLEMS)

Steps in solving maximum and minimum problems.
1. If needed, draw the figure and indicate the unknown
parts by variables.
2. Make the necessary representation for the variables
used.
3. Determine the quantity, the dependent variable, to be
maximized or minimized.
4. Find a formula relating the variables.
5. Express the dependent variable in step 3 in terms of a
single variable, the independent variable.
6. Differentiate the dependent variable in terms of the
independent variable.
7. Equate the derivative in step 6 to zero, for maximum or
minimum value, to find the critical numbers.
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.
Let x  be one part
120 - x  be the other part
M  be the product of one part and the square of the other part
Equation: M  120 - x x 2 dM
 0  240x - 3x 2  0
M'  120 - x 2x   x 2  1 dx
M'  240x - 2x 2  x 2 3x  0  x  0
3x80 - x   0  
M'  240x - 3x 2 80  x  0  x  80
d2M
Using the 2nd derivative test : 2
 240  6x
dx
d2M
when x  80; 2
 240  680      max imum
dx
d2M
when x  0; 2
 240  60      minimum
dx
answer : x  80 and 120 - x  40
Example 2
A box is to be made from a piece of cardboard
16in.x10in. by cutting equal squares out of the corners
and turning up the sides. Find the volume of the largest
box that can be made this way.

x x
x x x x
x x
10-2x 10
x x
x x x x
x x
16-2x

16
x

16-2x

V  LWh
Let L  length  16 - 2x
 16 - 2x 10  2x x 
W  width  10 - 2x 
 160 - 52x  4x 2 x 
h  height  x
V  160x - 52x 2  4x 3
dV
0  160 - 104x  12x 2  0 dV
dx  160 - 104x  12x 2
3x 2 - 26x  40  0 dx
x - 23x  20   0 V  16  2(2)10  2(2)2
20  (12)(6)(2)  144in3
x  2 and x   6.67
3
thus x  2 x must not be equal to 6.67
because 10-2x will be negative
Example 3
Find the area of the largest rectangle that can be
inscribed in a circle of a given radius, R = 10 inches.

2x

R R y
2y

x
Let L  length of the rectangle  2x
W  width of the rec tangle  2y
4x 2
A  LW  2x 2y  A'  0     0
2
4 100 x
100  x 2

A  4xy
 4x 2  4100  x 2 
since, R2  x 2  y 2 0
100  x 2

then y 2  R2  x 2  4x 2  400  4x 2  0
y  R2  x 2  8x 2  400
substituting y  R2  x 2 to A  4xy x 2  50
we have, A  4x R2  x 2 x  50  5 2 in.
A  4x 100 - x 2
therefore, A  4 50  100   50 
2

A'  4x
 2x 
  
100  x 2 4   4 50  50 
 2 100  x 
2
A  200 in.2
 4x 2
A'  4 100  x 2
100  x 2
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”.

10-y

H=10”
y

x
R=5”
Volume of a right cylinder : dV 
 100  40y  3y 2 
V  B h   R2 h dy 4
dV 
 0  100  40y  3y 2   0
V   x 2  y dy 4
x H y 3y 2  40y  100  0
but, 
R H 3y  10 y  10   0
x 10  y
 3y  10  0 y  10  0
5 10
10
510  y  10  y y   3.33" y  10"
x  3
10 2 extraneous
checking :

2
 10 y 
hence, V    y d2 V 
 2  2
  40  6y 
dy 4
 100  20y  y 
2

  y 10 d2 V    10  
if y  ,    40  6 
 4  3 dy 4 
2
 3 

V  100y  20y 2  y 3  d2 V
 5 max .
4 dy 2
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
Let P  the amount of fence needed to enclose the lots
A  area of the field
A
P  2L  4W and A  LW  W 
L
substituting W in P  2L  4W;
A
then, P  2L  4 
L dP 2L2  4 A
0  0
2L2  4 A dL L2

P
L 2L2  4 A but, A  LW
but A is cons tant, hence
2L2  4LW
dP L4L   2L2  4 A 1
 therefore,
1
L  2W or W  L
dL L2 2
dP 4L2  2L2  4 A 2L2  4 A
 2

dL L L2
Example 6
Find the most economical proportions for a quart can.

d
V   r 2h  Volume of cylinder
V
then, h  2
r
Solving for the Total Surface Area T  :
T  2 r 2  2 r h dT 2V
 0  4 r  2  0
dr r
 V 
 2 r  2 r 2 
2
2V
 r  4 r  2
r
V
T  2 r  2 
2
r 
3 2V V
 but, V   r 2h
r 4  2
 r 2h
dT   V1  r 
3

 4  r  2 2  2
dr  r  h d
dT 2V r since , r 
 4 r  2 2 2
dr r d h
  dh
2 2
Example 7
Find the proportion of the circular cylinder of largest
volume that can be inscribed in a given sphere.

r r
a
a h/2

C h 2a
Using Pythagorean Theorem; dV 3 2
 0   a  h  0
2

h
2
dh 4
a r  
2 2
3 2
2  a  h
2

2 4
h
a2  r 2  4 h2

4 h2  a2 substituting, a2  r 2 
2
3 4
h
r 2  a2  4  2 h2 
4 h  r  
2

3 4
but, V  Bh   r 2h
4 2 h2
 h 
2
h  r 
2
checking :
  a2  h 3 3
 4 h2 4 2 d2 V 3
 h3 h   r
2
  2h
V  a h
2 3 3 dh 2
4
4 3h2  h2 4 2 d2 V 3
 r
3 3   h  max.
dV 
  a2  3h2 
2
dh 2
2h2  4r 2
dh 4
h2  2r 2
dV 3
  a2   h2 h
dh 4 hr 2 or r 
2
EXERCISES:
1. What number exceeds its square by the maximum amount?
2. The sum of two numbers is “K”. find the minimum value of
the sum of their squares.
3. A rectangular field of given area is to be fenced off along the bank
of a river. If no fence is needed along the river, what are the
dimensions of the rectangle that will require the least amount of
fencing?
4. A Norman window consists of a rectangle surmounted by a
semicircle. What shape gives the most light for a given perimeter?
5. A cylindrical glass jar has a plastic top. If the plastic is half as
expensive as the glass per unit area, find the most economical
proportions for the glass.
6. Find the proportions of the circular cone of maximum volume
inscribed in a sphere.
7. A wall 8 feet high and 24.5 feet from a house. Find the shortest
ladder which will reach from the ground to the house when
leaning over the wall
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