finding the maximum area of an enclosed rectangle

Question

A farmer decides to enclose a rectangular garden using one side of a barn as one side of the rectangle. What is the maximum area the farmer can enclose with 100ft of fence? what should the dimensions of the garden be?

Answer 1

A more direct approach.

 

Let l be the dimension parallel to the barn's wall.

l + 2w = 100   Total length of fencing used.  Notice there's only one l.

l = 100 - 2w

 

A =  lw = (100-2w)(w) = -2w²+100w

This parabola describes area versus w, and it opens downward.

We already have it factored so we can easily find the roots for w when A = 0.

w = 0

100-2w = 0, w = 50.

The vertex of the parabola is halfway between the roots.

w_max = 25.

Plug this into A = -2w²+100w

A = (-2)(25²)+(100)(25)

   = (-2)(625) + 2500

   = -1250 + 2500

   = 1250

 

Answer 2

P=2l+2w

100=2l+2w

50=l+w --->50-w=l

A=lw

A=(50-w)(w)

A=50w-w^2

this is a parabola opening downwards whose maximum point is the vertex (h,k)

h=-b/2a

h=-50/-2=25 feet on each side will give you the maximum area

however, you don't need 25 feet where the barn is so add the 25 feet to the opposite side giving you 50 feet opposite the barn

the dimensions are 25 feet by 50 feet by 25 feet by "the barn"

A=lw

A=50*25=1250 square feet is the maximum area

notice that the rectangle is really 2 squares 25 x 25 side by side

the largest area is always a square so all you had to do was divide 100 by 4 to get 25 feet and take it from there