1. ## Finding area

A rectangular field is to be enclosed by a fence. An existing fence will form one side of the enclosure. The amount of new fence bought for the other three sides is 1200 feet. What is the maximum area of the enclosed field?

Thanks in advance for any help!

2. Originally Posted by live_laugh_luv27
A rectangular field is to be enclosed by a fence. An existing fence will form one side of the enclosure. The amount of new fence bought for the other three sides is 1200 feet. What is the maximum area of the enclosed field?

Thanks in advance for any help!
1) Let us say one side, x ft, is parallel to the existing fence.
So the two sides perpendicular to the existing fence are
(1200 -x)/2 ft each
Area, A, of the field is
A = x*(1200 -x)/2
A = (1200x -x^2)/2
A = -(1/2)(x^2) +600x .......a vertical parabola that opens downward ....because of the negative x^2, so its vertex is a maximum point.

If you know some Calculus,
dA/dx = -x +600
0 = -x +600
So, x = 600 ft for maximum A.
And so, (1200 -x)/2 = (1200 -600)/2 = 300 ft. each for max A.
Hence, maximum Area = (600)(300) = 180,000 sq.ft. -----answer.

If you don't know Calculus yet,
Using the properties of the parabola,
x at vertex = "-b /2a" = -600 /2(-1/2) = 600 ft
So, A at vertex = (-1/2)(600^2) +600(600) = 180,000 sq.ft. ----the max A, answer.

--------------------------
2) What if the sides perpendicular to the existing fence are x ft each?
Well, then, the side parallel to the existing fence will be (1200 -2x) ft.
So,
A = x(1200 -2x)
A = 1200x -2x^2
A = -2x^2 +1200x

That is again a vertical parabola that opens downward because of the negative x^2. The vertex is a maximum point.

x at vertex = "-b /2a" = -1200 /2(-2) = 300 ft each
A at vertex = -2(300^2) +1200(300) = 180,000 sq.ft. ----the max A, answer.

The same as in option #1 above.