Question

Isaac wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Isaac has 550 feet of fencing, you can find the dimensions that maximize the area of the enclosure.
a) Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn). Write an function for the area A of the enclosure in terms of w. (HINT first write two equations with w and l and A. Solve for l in one equation and substitute for l in the other).
b) What width w would maximize the area?
c) What is the maximum area?

Answer 1

Answer:

A:

Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn).

As only one side length will be considered so, we get the perimeter or the amount of fence material to form the rectangular enclosure as:

$550=l+2w$ or $l=550-2w$

We know that Area = length x width

So, $A=(550-2w)w$

$A=550w-2w^{2}$

B:

Area A is a function of w and it is a parabola that opens downwards so there is a maximum point.

We will take the vertex of the parabola as (h,k)

Here h is the maximizing number and k is the maximum area.

$h=\frac{-b}{2a}$

$\frac{-550}{2(-2)}$

$h=\frac{-550}{-4}$

h = 137.5

This is the length of all four sides except barn side. So, we will add 137.5 to 137.5 to get 275 feet

Now we have a rectangle that is 137.5 feet by 275 feet and 137.5 feet by barn. So, the maximum width will be 137.5 feet.

C:

And area will be = $137.5\imes275=37812.5$ square feet.

We can check the perimeter as : $137.5+275+137.5=550$ feet (GIVEN and also 3 sides are to be fenced.)