Question

Randy is building a fence at the side of his warehouse. He has 120 m of fencing and plans to use the

side of the warehouse as one side of the rectangular fenced area. What are the dimensions of the
maximum area Randy can enclose?​

Answer 1

The width (side perpedicular to the wall of the werehouse): x = 30 m

The lenght (side parallel to the wall of the werehouse): y = 60 m

Explanation:

x - the width of the fenced area

Randy has 120 m of fencing so for the lenght remain (120-2x) m of fencing:

y = 120 - 2x

Area of the fencing: A = x•y

A(x) = x•(120 - 2x)

A(x) = -2x² + 120x ← quadratic function

The maximum value of quadratic function occurs at:
`x=-\frac b{2a}`

a = -2 b = 120

`x=-\frac b{2a}=-(120)/(2\cdot(-2))=-(120)/(-4)=30`

y = 120 - 2•30 = 60