# What is the maximum area that the enclosure described below can contain?You have 32 m of fencing to enclosure a rectangular garden. One side of the garden will be the side of a barn. How do I solve...

What is the maximum area that the enclosure described below can contain?

You have 32 m of fencing to enclosure a rectangular garden. One side of the garden will be the side of a barn.

How do I solve this word problem?

hala718 | Certified Educator

Let x and y be the sides of the rectangle, Then the perimeter:

==> P = 2*x + 2*y

But one of the sides will not be fenced because it is open to the barn.

Then,we will remove one of the sides.

==> 32 = x + 2y

==> x = 32-2y.........(1)

Then the area :

A = x*y

= (32-2y)*y

= 32y - 2y^2

==> A = -2y^2 + 32y

We notice that we have a parabola facing down because y factor is negative, then we have a maximum point.

First we will find first derivative's zero.

A' = -4y + 32 = 0

==> y = 8

==> x= 32-2y = 32-16 = 16

==> x = 16

Then maximum area will be when the sides are 8 and 16.

==> A = 8*16 = 128 is the maximum area.

neela | Student

Since one side there is a barn, the fencing is only on 3sides of the rectangular garden.

So if x is one  of the two opposite sides, the then the 32 -2x is another ther side opposite to the barn.

So the area of the garden is maximum is x(32-2x) is maximum.

Let y = x*(32-2x)

y = 32x-2x^2

y/2 = 16-x^2

y/2 = a^2 - (a^2-16x+x^2)

y/2 = a^2 - (a-x)^2, when a = 8

y/2 = 8^2- (8-x)^2 is maximum when x=8 as (8-x)^2 > 0 for all real  x and  (8-x)^2   is zero when x = 8, making y/2 = 8^2 maximum . Or y = 2*8^2 = 128 is maximum value of y when x= 8.

So the measurements of the