Question

A rectangular lot is 125 yards long and 80 yards wide.
Give the length and width of another rectangular lot that has the same perimeter but a larger area.

Answer 1

Answer:

Step-by-step explanation:

perimeter=2(125+80)=2(205)=410 yards

let the length of new rectangle be x and y.

Perimeter=2(x+y)

2(x+y)=410

x+y=410/2=205 yards

y=205-x

area A=xy=x(205-x)=205x-x²

$\frac{dA}{dx} =205-2x\\\frac{dA}{dx}=0,gives~205-2x=0\\x=\frac{205}{2}\\\frac{d^2A}{dx^2}=-2<0 ~at~x=\frac{205}{2}$

A is maximum at x=205/2

y=205-205/2=(410-205)/2=205/2

so x=205/2=102.5 yards

y=205/2=102.5 yards