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

point a is the center of this circle.

the ratio of the lengths of (ef/bc) and (ad/ih) is 2: 1.

step-by-step explanation:

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer