Answer to Question #108765 in Calculus for mya

Question

A fence must be built to enclose a rectangular area of 45,000ft^2. Fencing material cost $1 per foot for the two sides facing north and south and $2 per foot for the other two sides. Find the cost of the least expensive fence.

Answer 1

Let $x=$ the width of rectangle in ft, $y=$ the length of rectangle in ft.

A fence must be built to enclose a rectangular area of 45,000 ft²

$$xy=45000=>y={45000 \over x}$$

The cost of the fence

$$C=2(2x)+1(2y)=4x+2y$$

Hence

$$C=C(x)=4x+2({45000 \over x}),\ x>0$$

Find the first derivative with respect to $x$

$$C'(x)=4-{90000 \over x^2}$$

Find the critical value(s)

$$C'(x)=0=>4-{90000 \over x^2}=0=>x_1=-150, x_2=150$$

First derivative test

If $x<-150,$ $C'(x)>0, C(x)$ increases.

If $-150<x<0,C'(x)<0,C(x)$ decreases.

If $0<x<150, C'(x)<0,C(x)$ decreases.

If $x>150,C'(x)>0, C(x)$ increases.

Since $x>0,$ then the function $C(x)$ has the absolute minimum at $x=150.$

So $y=\dfrac{45000}{150} =300$

The cost of the least expensive fence will be

$$C_{min}=2(2)(150)+1(2)(300)=\$1200$$