Answer to Question #163329 in Calculus for ary

Question

A rectangle plot of land is to be fenced off so that the area enclosed will be 400 ft2

. Let 

L

be the 

length of fencing needed and 

x

the length of one side of the rectangle. Show that

L =$2x$ + (800/x)

for 𝑥 > 0. What is the shortest possible length of such a fence?

Answer 1

Let x be one side of a rectangle and y the other. Then the area of a rectangle is, obviously, $S=xy=400\text{ ft}^2$ and thus $y = \frac{400}{x}$, x and y are both $>0$. Now the perimeter of this rectangle (and thus the total length of fence needed) is given by $L=2x+2y = 2x + 2\cdot \frac{400}{x}=2x+\frac{800}{x}$.

Now to find the minimum of a total length we will calculate the derivative of L with respect to x : $\frac{dL}{dx} =2-\frac{800}{x^2}$ and the zero of this derivative is $x_0 = 20$. Thus the minimal length is $L = 2\cdot 20+\frac{800}{20} = 80 \text{ ft}$.