A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Four 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?
The smaller dimension is ? feet

h = 120 ft;    w = 80 ft

A = 9600 ft^2

Step-by-step explanation: Let h and w be the dimensions of the playground. The area is given by:

A = h*w   (eq1)

The total amount of fence used is:

L = 2*h + 2*w + w    (eq2)    (an extra distance w beacuse of the division)

Solving for w:

w = L - 2/3*h = 480 - 2/3*h   (eq3)  Replacing this into the area eq:

We derive this and equal zero to find its maximum:

  Solving for h:

h = 120 ft.  Replacing this into eq3:

w = 80ft

Therefore the maximum area is:

A = 9600 ft^2

g(x) is the negative form of f(x)? because whenever its the negative form it reflects its self over the x axis. idk if this is right though.

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!