How do you find the maximum area of rectangle with 80ft perimeter?

1 Answer

Answer:

Algebraically, we can show that the maximum area of the rectangle is achieved when it is a square. In this case, the maximum area is #"400 ft"^2#

Explanation:

Since this is not in calculus, I'll provide a non-calculus answer.

We know the rectangle always has a perimeter of #80#, so #2l+2w=80#, which simplifies to be #l+w=40#.

We also know that the area of the rectangle is #A=lw#, but we can express this as a function of a single variable.

Use the perimeter expression #l+w=40# to say that #l=40-w#. Because this will always be true in the rectangle, we can substitute #40-w# for #l# in #A=lw#.

#A=lw#

#A=(40-w)w#

#A=-w^2+40w#

This quadratic function can be graphed, and the highest point will be the spot where the rectangle's area is maximized.

graph{-x^2+40x [-5, 45, -145, 460]}

The highest point, the vertex of the parabola, is #(20,400)#. When #w=20#, we also know that #l=20# (which forms a square) and the maximum area of the rectangle (square) is #"400 ft"^2#.