Arectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. seven hundred and eighty feet of fencing is used. find the dimensions of the playground that maximize the total enclosed area. what is the maximum area?

Step-by-step explanation:

Suppose the dimensions of the playground are x and y.

The total amount of the fence used is given and it is 780 ft. In terms of x and y this would be 3x+2y=780 (we add 3x because we want it to be cut in the middle). Therefore,  y= 780/2-3/2x. Now, the total area (A )to be fenced is

A=x*y= x*(390-3/2x)=-3/2 x^2+390x

Calculating the derivative of A and setting it equals to 0 to find the maximum

A'= -3x+390=0

This yields x=130.

Therefore y=780/2-3/2*130=195

Thus, the maximum area is 130*195=25,350ft^2

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.