A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Six hundredSix hundred and sixtyand sixty feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. What is the maximum area?

Dimensions: 165 feet by 110 feet.

Maximum Area =18,150 Square feet

Step-by-step explanation:

Let the dimension of the playground be x and y.

The rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground.

Let the side parallel to one side of the playground =x

Therefore, the total length of fencing =2x+x+2y

P=3x+2y

Six hundred and sixty feet of fencing is used.

We have: 3x+2y=660

3x=660-2y

Area of the Playground A=xy

We write the area in terms of y by substitution of x derived above.

We want to maximize the total enclosed area.

To do this, we first find the derivative of A(y).

Next, we solve A'(y) for its critical point.

Recall that:

Therefore:

Therefore, the dimensions of the playground that maximize the total enclosed area is 165 feet by 110 feet.

Maximum Area =165 X 110

=18,150 Square feet

first, we define the variables:

x: number of years after 1950

f (x): amount of vinyl sold.

then, with the variables defined, we have:

68594 vinyl records were sold in 1958 > f (8) = 68594

91299 vinyl records were sold in 1961 > f (11) = 91299

38720 vinyl records were sold in 1952 > f (2) = 38720

161743 vinyl records were sold in 1967 > f (17) = 161743

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step-by-step explanation: