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Like us? 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. $A=w(64-w)$ $A=64w-w^2$

We want to maximize A, so we need to graph A vs w, by completing the square. $A=-w^2+64w$ $A=-(w^2-64w +(64/2)^2) +(64/2)^2$ $A=-(w^2-64w+1024) +1024$ $A=-(w-32)^2 +1024$

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.