You are designing a rectangular and closure with a rectangular sections separated by parallel walls if you have 2500 feet of fencing what is the maximum area that can be enclosed

781250

Step-by-step explanation:

Let x = length of fenced side parallel to the side that borders the river

and y = length of each of the other two fenced sides  

Then, x + 2y = 2500

<=>  x = 2500- 2y

Area = xy = y(2500-2y)

            = - + 2500y

The graph of the area function is a parabola opening downward.

The maximum area occurs when y = -2500/[2(-2)] = 625

                                             x = 2500-2y = 1250

The maximum area that can be enclosed = xy = 1250*625 = 781250

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

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