Never seen a problem like this

Question

You have 180 yards of fencing to use on three sides of a garden. The fourth side is the tool
shed.
a. What is the maximum area that you can fence?
b. What are the dimensions of the garden?

Answer 1

2L +W =180  (I )

 

A = WL    / Area that needed to be maximized

 

Substitute  W from in terms of L from (1)

 

     w = 180 -2L

 

   A = ( 180 - 2L) L

 

  A= f( L ) =  - 2L^2 + 180 L  

 

         f( L ) is maximized at : L = -180/ -4 = 45

 

          w = 180 - 2L = 180 -90 = 90

 

           A rectangle with dimensions of 45* 90 = 4050 yrd² will be the answer.    

 

     

 

 

 

 

 

Answer 2

We can assign  aa length and width such that 2x+y=180 and we try to maximize the area xy. Since we know y=180-2x, this becomes 180x-2x^2. To maximize the area we need todifferentiate it and make sure its roots are either max or min poin so we get the right ones. For our area to be max we willl need to examine between points. More on this later..

 

Differentiated..180-4x. When we set it to zero we get x=45. So the dimensions are 45 by 90 with a total area of 405.

 

Hope that helps,

Deanna