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Section 3.6 Optimization Problems
Practice HW from Larson Textbook (not to hand in)
p. 203 # 3-15 odd, 19, 21, 27, 29, 31
Optimization Problems involve using calculus techniques to find the absolute maximum
and absolute minimum values (Steps on p. 198)
The following geometry formulas can sometimes be helpful.
Volume of a Cube: V  x 3 , where x is the length of each side of the cube.
Surface Area of a Cube: A  6x 2 , where x is the length of each side of the cube.
2
Example 1: Find the two numbers whose sum is 132 and product is a maximum.
Solution:
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3
Example 2: A farmer has 200 yards of fencing to fence in a rectangular pasture. One side
is next to a river and requires no fencing. Find the dimensions of the pasture that will
yield a maximum area.
Solution:
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4
1
Example 3: Find the point on the graph of f ( x)  x 2 that is closest to the point ( 2, ) .
2
Solution: In this problem, our goal is to minimize the distance between the ordered pair
1
(x, y) on the graph of y  f ( x)  x 2 and ( 2, ) .
2
To find the objective equation (the quantity that we want to minimize), we use the
distance formula.
Distance between th e two points
( x1 , y1 ) and ( x2 , y 2 )
 d  ( x2  x1 ) 2  ( y 2  y1 ) 2
Hence
Distance between th e two points
1
( x, y) and (2, )
2
1
 d  ( x  2) 2  ( y  ) 2
2
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To eliminate the radical, we can use the fact minimizing the distance d is equivalent to
minimizing the square of the distance, that is, minimizing d 2 . Hence, we write
1
D  d 2  ( x  2) 2  ( y  ) 2 .
2
Our goal is to minimize D. To do this, we must find the critical numbers. Before taking
the derivative, D must be in terms on one variable. We use the fact that we were given the
function y  f ( x)  x 2 . Hence, the equation for D becomes
1
D  ( x  2) 2  ( x 2  ) 2
2
Simplify, this equation becomes
1
1
D  ( x  2)( x  2)  ( x 2  )( x 2  )
2
2
or
D  x 2  2x  2x  4  x 4 
1 2 1 2 1
x  x 
2
2
4
(Use FOIL)
Simplifying gives
D  x 4  4x 
17
4
(Simplify. Note x 2 terms cancel and 4 
1 17

)
4 4
We next find the critical numbers. We set
dD
 4x3  4  0 .
dx
Solving for x gives
4x3  4
4x3 4

4
4
x3  1
x  3 1 1
(Continued on next page)
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Thus, the critical number is x = 1. The second derivative test, where
d 2D
2
 12x 2 , can
dx
be used to show this point does indeed produce a minimum. We see that
d 2D
2
 12(1)  12  0 , which is concave up. Since
2
d 2D
2
 12 x 2  0 for all values
dx
dx x1
of x and there are no other critical numbers, we have produced an absolute minimum.
The point x = 1 is the x-coordinate of the point on the graph of f ( x)  x 2 that is closest
1
to the point ( 2, ) . To find the y-coordinate of the point, we substitute into the original
2
function y  f ( x)  x 2 . Hence, we see that y  f (1)  (1) 2  1. Thus our answer is the
point
(1, 1)
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Example 4: A box with a square base and open top must have a volume of 32000 cm 3 .
Find the dimensions of the box that minimizes the amount of material used.
Solution:
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Example 5: A Norman window has a rectangle surmounted by a semicircle. If the
perimeter of the window is 30 ft, set up the objective and constraint equations and find
the dimensions of the window so that the greatest amount of light is admitted.
Solution: The basic shape of the window is given by the following diagram:
r = x/2
y
99
x
To maximize to greatest amount of light admitted, we must maximize the area of the
window. Let x be the width of the rectangle and y be the width. Note that the radius r of
x
the semicircle is half the width of the rectangle, that is, r  . To find the objective area
2
equation, we must find the area of the rectangle and the area of the semicircle. The area
of the rectangle is given as follows:
Area of Rectangle  (length)(wi dth)  xy
Recall that the area of a circle is given by the formula Ac   r 2 , where r is the circle’s
radius. Since we have a semicircle, the area is half of this formula. Thus, keeping in mind
x
that r  , we have
2
2
1
1  x
1 x2 1
2
Area of Semi - Circle   r      
  x2
2
2  2
2 4 8
Thus, the to get the total area, we add the area rectangle to the area of the semi-circle:
1
Total Area  A  (Area of Rectangle)  (Area of Semi - Circle)  xy   x 2
8
(Continued on next page)
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Hence, the objective equation for this problem is
1
A  xy   x 2
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To get the perimeter of the rectangle, we use the fact that the perimeter around the
window is 30 ft. To get the perimeter, we must sum the length of sides around the
rectangle and the semicircle. The perimeter of the rectangle is
Perimeter of Rectangle  x  2 y
To get the perimeter around the rectangle, recall that the formula for the circumference of
a circle is given by C  2 r , where r is the circle’s radius. Since we have a semicircle,
x
the circumference is half of this formula. Thus, keeping in mind that r  , we have
2
Perimeter of Semi - Circle 
1
x
1
2 r   r   ( )   x
2
2
2
Thus, we have
1
Total Perimeter  30  (Perimeter of Rectangle)  (Perimeter of Semi - Circle)  x  2 y   x
2
Thus, the constraint equation is given by
1
x  2 y   x  30
2
Thus, the objective and constraint equations are given by the following formulas:
1
Objective: A  xy   x 2 ,
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1
Constraint: x  2 y   x  30
2
To maximize the objective equation, we need to convert it to function of one variable.
The easiest way to do this is to solve the constraint equation for y. We do that as follows:
1
x  2 y   x  30
2
1
2 y  30  x   x
2
1
1
y  (30  x   x)
2
2
1
1
y  15  x   x
2
4
(Continued on next page)
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If we substitute y  15 
1
1
1
x   x into the objective equation A  xy   x 2 , we
2
4
8
obtain
1
1
1
1
A  xy   x 2  x(15  x   x)   x 2
8
2
4
8
 15 x 
1 2 1
1
x   x2   x2
2
4
8
 15 x 
1 2 2
1
x   x2   x2
2
8
8
 15 x 
1 2 1
x   x2
2
8
Now, we take the objective equation A  15 x 
1 2 1
x   x 2 we compute the derivative
2
8
1
1
A  15  x   (2 x)  15  x   x , set it equal to 0, and solve for x to find the critical
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4
numbers:
1
15  x   x  0
4
1
15  x   x
4
1
x   x  15
4
1
x(1   )  15
4
x
15
15
15
60



 8.4 ft
1
4 
4
4
1 

4
4 4
4
1
Since A  1    0 for all values of x, this value produces a maximum. To find the
4
60
length y of the window, we substitute x 
back into the constraint relationship
4
(Continued on next page)
12
y  15 
1
1
x   x . This gives
2
4
y  15 
1
1
1 60
1
60
30
15
x   x  15 
 
 15 

 4.2 ft
2
4
2 (4   ) 4 (4   )
(4   ) (4   )
Thus, the dimensions
x
60
 8.4 ft
4
and
y  15 
30
15

 4.2 ft
(4   ) (4   )
will maximize the area of the window and yield the most light.
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