Fencing Math Equation Problems

Question

1. A homeowner has just enough money to purchase 200 feet of fencing for his backyard. He wants to use these 200 feet to enclose the greatest possible area. Make sure that the width of the enclosure is the shorter side, and label it x. Express the area of the enclosure as a function of the width. Find the maximum area of the enclosure using the function. What are the dimensions? What can you conclude?
2. The homeowner could enclose even more area if he used the back of his house as one side. Write an equation for this new area. Find the maximum area of the enclosure using the function. What are the dimensions? What can you conclude?

Answer 1

I am surprised that this problem was labeled under Algebra 2 or geometry.

This is a pretty classic Applied Optimization problem in Calculus 1.

Anyhow, the perimeter formula would be 2x +2y = perimeter, where x is the width and y the length.

P = 200 ft = 2x + 2y

Area = xy

Using the perimeter solve for y so you can replace it in the area equation will express that function in terms of x only (width).

200 = 2x + 2y =

y = (200 -2x)/2y

y = 100 -x

Substitute

A = x(100-x) = 100x -x²

Now take the derivative of A = 100x - x²

dA/dx = 100 -2x

Set the derivative to 0

0 = 100 - 2x

x = 50 ft.

The width is optimized at 50 ft. Take that and plug back into the original perimeter function and solve for y.

200 = 2(50) + 2y

y = 50 ft

For the 2nd part of the problem you change the perimeter equation to P = 2x + y

200 = 2x + y

y = 200 - 2x

A = xy = x(200-2x) = 200x - 2x²

and take the derivative and set to zero

dA/dx = 200 - 4x

dA/dx = 0 = 200 - 4x

x = 50ft

Plug back into the perimeter equation. y = 100 ft