Optimization Problems
1
Find 2 numbers whose difference is 100 and whose product is a
minimum.
2.
A rectangular field is to be fenced off along the bank of a river
and no fence is required along the river. If the material for the
fence cost $4 per foot for the two ends and $6 for the side
parallel to the river, find the dimensions of the largest possible
area that can be enclosed with $1800 worth of fence.
3.
If  cm of material is available to make a box with a
square base and an open top, find the largest possible volume
of the box.
4.
Find the point or points on the curves below which are closest
to the origin.
  
  
a)  
b)  
c)     


For part c), just get the x value.
5.
A cardboard box manufacturer wishes to make open boxes from
pieces of cardboard 12 inches square by cutting equal squares
from the four corners and turning up the sides. Find the length
of the side of the square to be cut out in order to obtain a box
of the largest possible volume.
6.
For the cost and demand function given below, find the
production level that will maximize revenue. Find the production
level that will maximize profit. What price will result in the
maximum profit?
 ab        ab    
7.
Find the dimensions of the rectangle of largest area that can be
inscribed in a circle of radius 
Solutions
1.
Let  and  be the 2 numbers. Then      and
    
Their product is     a  b     
To find where  is a minimum, find where    
        when    
Mark off   on a number line and make some test values
in   ab    
P is Decr
P'(-60)= - 20 < 0
P is Incr
-50
P'(0)=100 > 0
This analysis proves that   produces the minimum product.
If     then     a  b   Therefore the 2
numbers are  and  
2.
x
x
y
  
Since the cost per foot for each end is $4, the cost for x ft is 
and the cost for the y ft is  The total cost is
             
  


   .




The area is     Œ  •     



The maximum area will occur when       



           ft and



    ab   ft.

You can see that    produces a maximum A by
selecting test values and substituting into  
0 ----------- 112.5 -------  
  
  is incr. on a b and decr. on a b 
So the side parallel to the river is 150 ft and each of the other
2 sides is 112.5 ft.
3.
y
x
x
     we must get rid of x or y. We can do this
by using the fact that the total amount of material
in cm should be 1200 if the volume is maximized.
To get the amount of material (in cm ) note that there is a bottom
of area  and 4 sides, each with area   Therefore the total
  
area is        
.

  
  

  

    





                   


You can verify by using test values in   that 20 is where
the maximum volume occurs and

   
the maximum volume is  



 
  cm .


4.
  
a) The graph of  
is shown below with an

arbitrary point marked off in the 1st quadrant.
(0,9)
(x,y)
The distance squared from a b to a b is   
  


Let      and  
     


      
Therefore        where   9 
The distance from a b to a b will be a minimum if  ,
the distance squared, is a minimum. This could occur when
    or at the endpoint   
             
 ab         and  ab        
and the minimum  occurs when   
If    then
    ab       
The 2 points which are closest are a   b
  
has a similar graph where the highest

point is a b
b)  
(0,2)
(x,y)
  
           

           where   .
 will be a minimum when     or when   
           This value is not in the

interval      Therefore the minimum must occur
when    which means that   
Therefore the closest point is the vertex a b
c)
y = 4x+7
(x,y)
The distance squared from a b to a b is    
If      and      then
    a  b .
Again, the distance will be a minimum if  , the distance
squared is a minimum.
     a  b       


        
 



5.
x
x
x
x
12 - 2x
x
x
x
x
12 inches
The volume of the box is   a  b  
a     b      
         a    b  a  ba  b
    when    and 6.
------2-------6-----inc dec inc
Making some test value in   we can see that  is
increasing from 0 to 2 and decreasing from 2 to 6.
 can't be 6 or more. Therefore the maximum volume
will occur when    inches and the maximum volume
is a  b    in3 .
6.
7.
 ab        ab    

The revenue is given by  ab  ab   



The revenue is a maximum when   ab   


         .

Profit  ab   ab   ab   
     




  

    
  .
 

Profit is a maximum when



  ab   
        
 



The price that produces maximum profit is

   
 $11.33 .

Let  and  be as shown in the diagram below where we have
a rectangle inscribed in a circle of radius  The rectangle has
length  and width 
r
y
x
Then the area of the rectangle is
   where            
Note that  will be a maximum when  is a maximum.
      a   b      

      ˆ   ‰  

when    or when     
We don't want    so look at      




     



È


.
È








Then          



È

 
 


È

Therefore the dimensions of the rectangle are
   which is È by È . It's a square.