Answer to Question #274821 in Calculus for Olivia

Question

A rectangular lot adjacent to a highway is to be enclosed by a fence. If the fencing costs $2.50 per foot

along the highway and $1.5 per foot on the other sides, find the dimensions of the largest lot that can

be fenced off for $720.

Answer 1

Let $x=$ the length of a fence along the highway, $y=$ the width of a rectangular lot.

Then

$$2.5x+1.5(x+2y)=720$$

$$x=\dfrac{720-3y}{4}$$

The area of the rectangle is

$$A=xy$$

Substitute

$$A=A(y)=(\dfrac{720-3y}{4})y, 0<y<240$$

Find the first derivative wit respect to $y$

$$A'(y)=((\dfrac{720-3y}{4})y)'=180-1.5y$$

Find the critical number(s)

$$A'(y)=0=>180-1.5y=0$$

$$y=120$$

Critical number:$120.$

$$A(0)=0$$

$$A(240)=0$$

$$A(120)=(\dfrac{720-3(120)}{4})(120)=10800$$

The area has the absolute maximum with value of $10800$ for $0<y<240$

at $y=120.$

$$x=\dfrac{720-3(120)}{4}=90$$

The length of a fence along the highway is $90 \ ft,$ the width of a rectangular lot is $120\ ft.$