2.13 Topics: Optimization problems Solutions
1. Find two positive numbers such that the product is 147 and the sum of the first number plus three times the second number is a
minimum.
Let the two positive numbers be x and y. PRIMARY EQUATION: m  x  3 y SECONDARY EQUATION: xy  147
xy  147 
y
147
x
441
 147 
m  x  3y  m  x  3
  m  x
x
 x 
x   0,  
441 x 2  441  x  21 x  21


x2
x2
x2
The only critical number that is in the feasible domain is x  21
Note: The second derivative test shows this is indeed a minimum
882 882
m  0  3  3  m  21  
x
x
m  1 
Thus the answer to this question is the two numbers are x  21
y
147 147

 7
x
21
2. Find two positive numbers such that the sum of the first number and twice the second number is 108 and the product is a
maximum.
Let the two positive numbers be x and y. PRIMARY EQUATION: P  xy SECONDARY EQUATION: x  2 y  108
x  2 y  108  x  108  2 y
P  xy  P  108  2 y  y  P  108 y  2 y 2
y   0,54 
P  108  4 y  4  27  y 
The only critical number, 27, is indeed in the feasible domain and
the second derivative test shows it is a maximum.
P  4  P  27   
So the two numbers we are looking for are y  27
x  108  2 y  108  2  27   54
3. Find the length and width of a rectangle with perimeter 80 meters whose area is a maximum.
Let L and W be the length and width of the rectangle (see diagram)
PRIMARY EQUATION: A  LW SECONDARY EQUATION: 2L  2W  80
2 L  2W  80  L  W  40  L  40  W
W
A  LW  A   40  W W  A  40W  W 2
W   0, 40 
A  40  2W  2  20  W 
The only critical number, 20, is in the feasible domain and
the second derivative test shows that it is a maximum.
A  2 
A  20   
So the Length of the rectangle is 20 meters and the Width of the rectangle is 20 meters.
(the rectangle is a square).
L
4. Find the length and width of a rectangle with area 32 ft 2 and whose perimeter is a minimum.
Let L and W be the length and width of the rectangle (see diagram)
PRIMARY EQUATION: P  2L  2W SECONDARY EQUATION: LW  32
32
LW  32  L 
W
64
 32 
P  2 L  2W  P  2    2W  P 
 2W
W
W 
W
W   0,  
L
2
64
2W 2  64 2 W  32 


2

P


W2
W2
W2
The only critical number that is in the feasible domain is W  32  4 2
and the second derivative test shows that there is indeed a minimum there.
128
P   3  P  4 2  
W
So the length and width that will give the minimum perimeter will be
32
32
W= 4 2 ft
L

= 4 2 ft
W 4 2
P  


 1
5. Find the point on the graph of f  x   x 2 that is closest to the point  2, 
 2
 1
Let  x, y  be the point on the graph of f  x   x 2 that is closest to the point  2, 
 2
 1
We want to “minimize” the distance between  x, y  and  2,  so the distance formula will
 2
be our primary equation. Also note that a reasonable feasible domain will be
y
f(x)=x^2
Series 1
6.5
x  0, 2 look at the picture to see why this is reasonable.
6
5.5
Primary Equation: d 
 x  2
2
1

 y 
2

2
5
Secondary Equation: y  x 2
4.5
4
3.5
2
1
17

  x2    d  x4  4x 
2
4

because d is smallest when the expression inside the radical is smallest we only
d
 x  2
3
2
2.5
2
1.5
1
17
need to find the critical numbers of f  x   x  4 x 
4
3
3
2

f  x   4 x  4  4  x  1  4  x  1  x  x  1
4
the only REAL critical number is at x = 1 (note this is in our feasible region  0, 2  )
Using the Extreme Value Theorem...
17
5
7
d 1 
d  2 
2
2
2
So we see the distance is minimized when x = 1. So the closest point on y  x 2
d 0 
 1
to the point  2,  is the point 1,1 .
 2
0.5
x
-2
-1.5
-1
-0.5
0.5
-0.5
1
1.5
2
6. Find the point on the graph of f  x   x that is closest to the point  4,0
We will use a similar approach on this problem that we did on the previous one.
Let  x, y  be the point on the function f  x   x that is closest to the point
f(x)=x^(.5)
Series 1
 4,0
.
4
 x  4
Primary Equation: d 
d
y
4.5
 x  4
2
  y  0
2
2
  y  0
 d
2
 x  4
Secondary Equation: y  x
2


x 0

3.5
3
2.5
2
 d  x 2  7 x  16
2
A feasible domain for x will be x   0,5 see the diagram to convince yourself of this.
1.5
Minimizing the radicand will be the same as minimizing d
(see last problem explanation).
1
0.5
x
-0.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
-0.5
f  x   x 2  7 x  16
f   x   2 x  7 So the only critical number is x 
7
2
Note: this is in our feasible domain.
d 0  4
7
d    1.9365
2
d  5  6
So by the Exteme Value Theorem we can see the minimum distance will occur when x =
7
2
7 7 
So the point on the graph of y  x that is closest to the point  4, 0  will be  ,

2 2
7. A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 245,000 m 2 in order to provide enough
grass for the herd. What dimensions will require the least amount of fencing if no fencing is required along the river?
PRIMARY EQUATION: P  x  2 y
SECONDARY EQUATION: xy  245000
RIVER
245000
y
245000
P  x  2y  P 
 2y
y
xy  245000  x 
Y
Y
X
A feasible domain for y is y   0,   .
P  
P 
245000
245000  2 y 2
 2  P 
2
y
y2
 P 
2  y 2  122500 
y2
2  y  350  y  350 
y2
The only critical number that is in our feasible domain is 350. The second
derivative test shows this to be a maximum.
P  
490000
y3
 P   350   
So the dimensions of the rectangular pasture should be y = 350m
and x 
245000 245000

 700m
y
350
8. A rectangle is bounded by the x-axis and the semicircle y  25  x 2 . What length and width should the rectangle have so that its
area is a maximum?
Primary Equation: A  2 xy
A  2 xy 
5
 x, y 
Secondary Equation: y  25  x 2
A  2 x 25  x 2
A feasible domain for x will be x   0,5  Look at the picture to see why.
A  2 x 25  x 2
1
1
1


 1

A  2  x   25  x 2  2  2 x    25  x 2  2 1  2  25  x 2  2  x 2   25  x 2 
 2


-5
5
A 

2  2 x 2  25 
25  x 2
25
5
5 2


 3.54
2
2
2
The first derivative test will show this yeilds a maximum (note in this case it is
easier to do the first derivative test than to do the second derivative test).
The only critical number in our feasible domain is x 
 5 2
 0,

2 

test #
1
sign of A
+
5 2 
,5 

 2

4

So the length of the rectangle (2x) should be 5 2 and the width (y) should be
2
2
5 2 
5 2 
100 50
50 5 2
y  25  x  25  



  25  
 
4
4
4
2
 2 
 2 
2
9. The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of the triangle and the square that
produce a minimum area.
Perimeter of the square = 4x Perimeter of the equilateral triangle = 3y
Primary Equation: A = area of the square + area of the triangle
(the height of the triangle can be found by dropping a perpendicular line
from the upper vertex and using the Pythagorean theorem you will find
height =
x
3
y .)
2
y
x
1  3 
Secondary Equation: 4 x  3 y  10
y
y
2  2 
10  3 y
4 x  3 y  10  x 
4
2
2
10  3 y 
3 y2
1  3 
 10  3 y  1  3 


A

y
y

y   A  
A  x 2  y 



4
16
2  2 
 4  2  2 
 10 
A feasible domain for y will be y  0, 
 3
Primary Equation: A  x 2 
A 
30
3
1
10  3 y  3  y setting A  0 we can find the only critical number y 
2
8
94 3
 30 
 10  25 3
 4.81
A
  2.719 A   
9
 3
94 3 
So by the extreme value theorem we can conclude that the area will be a minimum when the
A 0 
25
 6.25
4
sides of the equilateral triangle are each y 
30
94 3
and the sides of the square are each
 30 
90
90
94 3
10  3 
10 
 10 
10 9  4 3  90
10 3
10  3 y
94 3 
3
4

9
3
4

9

1






x
4
4
4
4
4 3
9
94 3
4 94 3
1




y
y