Calculus (6th edition) by James Stewart
Section 4.7- Optimization Problems
3.
Find two positive numbers whose product is 100 and whose sum
is a minimum.
Let  and  be the 2 numbers. Since the product of these numbers

is 100,      
  

The sum is given by the function     
where     
The minimum value of  will occur when    


                



        (forget about -10).

The minimum of  will occur when      
 

Ans: The 2 numbers, which are identical, are 10 and 10.
11.
A farmer wants to fence an area of 1.5 million square feet in a
rectangular field and then divide it in half with a fence parallel to one
of the sides of the rectangle. How can he do this so as to minimize
the cost of the fence?
The cost is minimum if the amount of fencing is a minimum.
y
x
The amount of fencing is     don't forget the fencing
in the middle.
Let       We must get  in terms of one variable.
We know that area is  and   ,  
Hence  
  
    

    ˆ   ‰      
where      
If  has a minimum it must occur at a critical value So
find where    
  
          
 

  

      

  
Ê
  feet





  feet.


To minimize the cost, the rectangle should be 1000 feet by
1500 feet.
19.
Find the points on the ellipses      that are farthest away
from the point a b
The ellipse is shown below. Note that the x and y intercepts of
the ellipse are a  b and a   b
(x, y)
(1, 0)
Take an arbitrary point on the ellipse a b The distance from
a b to a b is È      The distance is a maximum
if and only if the square of the distance is a maximum. So we can
maximize a  b   
Let   a  b    we must get rid of  or  
It is better to substitute for  and keep the 
Since            
  a  b     where      .
 will be a maximum when    or    
             

when     or    

Now is  a maximum at      or   
a  b  a  b    a  b  
ab  
a  b        a  b  
We see that  is maximum when     
        ab   
    

The points which are farthest away are Œ     •

33.
A piece of wire 10 m long is cut into two pieces. One piece is bent
into a square and the other is bent into an equilateral triangle. How
should the wire be cut so that the total area enclosed is
a) A maximum?
10 - x
x
If  is the length of on piece that is cut, then the other piece
has length   
If the piece of length    is bent into a square then the
length of a side is a  b
Then the length of the side of the triangle is 
h
x/3 60

 È
È
 ° 





The base of the triangle has length 

The area of the triangle is abab 

È
È
   
 
Š ‹


 


Since ab   °  
The total area is  
a  b
 È
a  b







 È

where      
Assume that it is possible that    (no triangle)
or  (no square).
The maximum area will occur when   ,    or   

È



 
 a  b 
    È  



 
 È

 Ž 

    
•





 


ab 
  m

   m
 È
ab 
   m

The maximum total area is 6.25 m and this occurs when   
and there is no triangle, only the square.
b) A minimum?
The minimum will occur when     or 10.
From the computation above, the minimum area is 2.72 m .
This occurs when use  m for the triangle and
a  b   m for the square.