Question 71238
Let x = the width and y = length of the patio. The perimeter of the patio can be represented by {{{P=2x+2y=300}}}. The area by {{{A=xy}}}. See so far?.

Solve 2x+2y=300 for y and sub into the area formula.

{{{y=(300-2x)/2=150-x}}}

Now, sub into area, xy:

{{{x(150-x)=-x^2+150x}}}

You probably know the x-coordiante of the vertex is given by {{{-b/(2a)}}}

a=-1 and b=150

{{{-150/(2(-1))=75}}}

Therefore, the width x, is 75. 

Subbing gives us y=75 also.

Therefore, max area is achieved when the width is 75 and the length is 75.

It can be shown through calculus that the max area is achieved when the area is a square. Which you have. 75X75 square patio.