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Paul has 400 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum​ area?

A rectangle that maximizes the enclosed area has a length of _ nothing yards and a width of _ nothing yards.

The maximum area is _ nothing square yards.

Respuesta :

irspow
P=2(x+y)  perimeter equals two time the sum of x and y dimensions...

We are told that P=400 so

2(x+y)=400

x+y=200

x=200-y...

Now area is:

A=xy, using x found above we have:

A=(200-y)y

A=200y-y^2

The rate of change of the area is dA/dy

dA/dy=200-2y

The maximum area occurs when the rate of change is zero or dA/dy=0 so

200-2y=0

2y=200

y=x=100yd

So the maximum area enclosed is when y=x=100 which is a perfect square (as is always the case for a four sided enclosure with a given amount of material)

A=100^2=10000 yd^2
MrRoyal

The area of a shape is the amount of space it occupies.

  • A rectangle that maximizes the enclosed area has a length of 100 yards and a width of 100 yards.
  • The maximum area is 10000 square yards.

The perimeter is given as:

[tex]P = 400[/tex]

The perimeter of the fence is calculated as:

[tex]P = 2 \times (L + W)[/tex]

Where

L and W are the length and width of the fence

So, we have:

[tex]2 \times (L + W) = 400[/tex]

Divide both sides by 2

[tex](L + W) = 200[/tex]

Make L the subject

[tex]L = 200 -W[/tex]

The area of a rectangle is:

[tex]A =L \times W[/tex]

Substitute [tex]L = 200 -W[/tex]

[tex]A =(200 -W) \times W[/tex]

[tex]A =200W -W^2[/tex]

Differentiate, and set to 0

[tex]A' = 200 - 2W[/tex]

Set to 0

[tex]200 - 2W = 0[/tex]

Collect to 0

[tex]2W = 200[/tex]

Divide by 2

[tex]W = 100[/tex]

Recall that:

[tex]L = 200 -W[/tex]

[tex]L = 200 -100[/tex]

[tex]L =100[/tex]

The area of the fence is:

[tex]Area = 100 \times 100[/tex]

[tex]Area = 10000[/tex]

Hence,

The dimension that maximizes the area is 100 yards by 100 yards, and the maximum area is 10000 square yards

Read more about areas at:

https://brainly.com/question/11906003

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