Question

A local community center built a rectangular basketball court using 100 feet of fence. They used part of one side of the building as one length of the rectangular court. They choose to maximize the area using the 100 feet of fencing. The community center also wanted to include a small playground enclosed by a fence using 60 feet of fencing. Again, they utilized the side of the building as one length of the rectangular playground. The length of the playground is 30 feet less than the length of the basketball court. The width of the playground was 5 feet less than the width of the basketball court. How much larger is the basketball court than the playground

Answer 1

Answer:

Maximum Area of the basketball court = 1250 ft²

Maximum area of the playground = 400 ft²

The basketball is 3.125 times larger than than the playground.

Step-by-step explanation:

The diagram representing the description is attached to this solution.

The big rectangle represents the basketball court.

The small rectangle represents the playground.

Let the length and width of the basketball court be y and x respectively.

The length and width of the small playground will be (y-30) and (x-5) respectively

The Area of the basketball court is

A(x, y) = xy

But, 2x + y = 100

We can maximize the Area by turning the two variable function into a one variable function and substituting for y in the Area equation.

2x + y = 100

y = 100 - 2x

A(x, y) = xy

A(x) = x(100 - 2x) = 100x - 2x²

A(x) = 100x - 2x²

At maximum point, (dA/dx) = 0 and (d²A/dx²) < 0, that is, negative.

(dA/dx) = 100 - 4x = 0

(d²A/dx²) = -4 < 0

Hence, it's a maximum point.

At maximum point,

(dA/dx) = 100 - 4x = 0

100 - 4x = 0

4x = 100

x = 25 ft

y = 100 - 2x = 100 - 2(25) = 100 - 50 = 50 ft

Hence, the length and width of the basketball court that maximizes the Area are 50 ft and 25 ft respectively.

For the small playground,

The length = (y - 30) = (50 - 30) = 20 ft

The width = (x - 5) = (25 - 5) = 20 ft

Maximum Area of the basketball court = xy = (50)(25) = 1250 ft²

Maximum area of the playground = (y-30)(x-5) = (20)(20) = 400 ft²

Ratio of the Areas = (1250/400) = 3.125

The basketball is 3.125 times larger than than the playground

Hope this Helps!!!