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.

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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.