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Lemur [1.5K]
3 years ago
14

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

undred and eighty feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. What is the maximum​ area?
Mathematics
1 answer:
Arisa [49]3 years ago
8 0

Answer:

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

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Solution:

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To find the solution of the inequality:

3 x+8 \geq 11

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The graph of the solution is attached below.

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Answer:

D

Step-by-step explanation:

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