Question

A family wants to fence a rectangular play area alongside the wall of their house. the wall of their house bounds one side of the play area. if they want the play area to be exactly 25002500ft22, what is the least amount of fencing needed to make this? round your answer to the nearest two decimal places.

Answer 1

The minimum amount of fencing would be 150 ft.

When maximizing area and minimizing perimeter, we want the dimensions to be as close to equal as possible.  We start out taking the square root of the area, 2500:

√2500 = 50

Since the square root is a whole number, we can use 50 for both the length and width of the play area.  Since the house makes up one wall of the area, we need 3 sides that are 50 ft each:

3(50) = 150 ft.

Answer 2

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

The least amount of fencing is 141.42 ft

The area is given as:

$\mathbf{A = 2500}$

Let the dimensions be x and y.

So, we have:

$\mathbf{A = xy}$

$\mathbf{P = 2x + y}$

Substitute 2500 for A in $\mathbf{A = xy}$

$\mathbf{xy = 2500}$

Make y the subject

$\mathbf{y= \frac{2500}{x}}$

$\mathbf{P = 2x + y}$ becomes

$\mathbf{P = 2x + \frac{2500}{x}}$

Rewrite as:

$\mathbf{P = 2x + 2500x^{-1}}$

Differentiate

$\mathbf{P' = 2 - 2500x^{-2}}$

Set to 0

$\mathbf{2 - 2500x^{-2} = 0}$

Subtract 2 from both sides

$\mathbf{ - 2500x^{-2} = -2}$

Make x^2 the subject

$\mathbf{x^2 = \frac{2500}{2}}$

$\mathbf{x^2 = 1250}$

Take square roots

$\mathbf{x = 35.36}$

Recall that: $\mathbf{P = 2x + \frac{2500}{x}}$

So, we have:

$\mathbf{P = 2 \times 35.36 + \frac{2500}{35.36}}$

$\mathbf{P = 141.42}$

Hence, the least amount of fencing is 141.42 ft

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