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Max/Min Rectangle Problem, Maximize the Area

Q: You have 128 feet of fencing to fence a rectangular area. Find the largest possible area to enclose this amount.

A: The perimeter of the space will be 128=2w+2L, where w=width and L=length

Isolate for L: L=64-w

The formula for the area of the rectangle is A=w*L. Substitute L=64-w into the area formula so that we only have one variable, w.


We want to maximize A, so we need to graph A vs w, by completing the square.

A=-(w^2-64w +(64/2)^2) +(64/2)^2
A=-(w^2-64w+1024) +1024
A=-(w-32)^2 +1024

Graph this to see that the maximum area of 1024 ft^2 occurs when w=32 ft. This makes sense, as it is a 32 x 32 square. A square gives the Max area.

Here’s a link to a graph of the function. Note that area is the y axis, and width is the x axis.