math is fun ....................................

Question

Austin 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 Austin has 550 feet of fencing, you can find the dimensions that maximize the area of the enclosure.

Answer 1

We need an equation for the area with just one dimension. The equation for area is A = L x W

But the total perimeter (which is 550) is W x 2L or (vice versa)

So, we have 550 = W + 2L

Now we can solve for W in terms of L and plug that into the equation for area:

550 = W + 2L

W = (550 - 2L)

**I put parenthesis to show what's being replaced/plugged in**

A = L x (W)

A = L x (550 - 2L)

Now distribute:

A = 550L - 2L²

As a function, this is a parabola. The negative in front of the squared term means the parabola points downward, thus having a maximum.

We can solve for the maximum by finding the vertex. The x coordinate of the vertex is equal to -b/2a.

I'm going to rewrite the equation with y and x instead of A and L, and I'm going to place the squared term in front so it looks more familiar:

y = -2x² + 550x

b is 550

a is -2

x = -(b)/2(a) = -(550)/2(-2) = 550/4 = 137.5

So, at the vertex/maximum, x = 137.5

We said x was L or length, so we can easily solve for W now:

550 = W + 2L

550 = W + 2(137.5)

W = 275

So you have two sides which are 137.5ft each and one side which is 275ft.

**If you needed to solve for the area, you would simply plug 137.5 in for x in the parabolic equation and solve for y:

y = -2(137.5)² + 550(137.5)

y = -37812.5 + 75625 = 37812.5 ft²

You can also check to see if it is the max by plugging in L and W values into the area equation that are slightly higher/lower. For example:

550 = (274) + 2(138)

274 x 138 = 37812 ft²

and

550 = (276) + 2(137)

(276) x (137) = 37812 ft²

As you can see, trying both a slightly lower value for W and then a slightly higher value for W gives us a slightly smaller area.