Question

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?

Answer 1

Answer:

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

$x=\dfrac{660-2y}{3}$

Area of the Playground A=xy

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

$A(y)=y\left(\dfrac{660-2y}{3}\right )\\A(y)=\dfrac{660y-2y^2}{3}$

We want to maximize the total enclosed area.

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

$A'(y)=\dfrac{660-4y}{3}$

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

$A'(y)=\dfrac{660-4y}{3}=0\\660-4y=0\\660=4y\\y=660 \div 4\\y=165 feet$

Recall that: $x=\dfrac{660-2y}{3}$

Therefore:

$x=\dfrac{660-2(165)}{3}=\dfrac{660-330}{3}=\dfrac{330}{3}\\x=110 feet$

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