# If the area of a rectangle is 32 feet, what is the perimeter?

## Question:

If the area of a rectangle is 32 feet, what is the perimeter?

## Rectangle: Area to Perimeter Conversion:

Let's say that we are provided with a rectangle of two unknown sides: length {eq}x {/eq} and width {eq}y {/eq}. The area of such a triangle is given by {eq}A(x, y) = xy {/eq}. Additionally, we can express the perimeter as the sum of two lengths and two widths: {eq}P(x, y) = 2x + 2y = 2(x + y) {/eq}

Knowing the area of a rectangle, we can express one of the variables in terms of the other and obtain the perimeter function of a rectangle.

Given the dimensions of the rectangle by {eq}x {/eq} and {eq}y {/eq}, we can obtain the area function:

{eq}A(x, y) = xy = 32~\rm{ft^2} {/eq}

Expressing {eq}y {/eq} in terms of {eq}x {/eq}, we obtain:

{eq}y = \frac{32}{x} {/eq}

The perimeter of this rectangle is:

{eq}P(x, y) = 2(x + y)\\ P(x) = 2(x + \frac{32}{x}) {/eq}

Since we do not have an additional equation to solve for x, we cannot evaluate the actual perimeter and it depends on x. For example, given {eq}x = 1 {/eq}, we obtain a perimeter of {eq}P(1) = 2(1 + \frac{32}{1}) = 66~\rm{ft} {/eq}

In a general case, the perimeter of the described set of triangles is calculated by {eq}\boxed{P(x) = 2(1 + \frac{32}{x})\\ x > 0} {/eq}