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

Question

Nathan Axel · Accepted 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#

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#.

Ethan Adkins · Answer

undefined To find the maximum area of a rectangle with an 80-foot perimeter, you can use the formula for the perimeter of a rectangle, which is $ P = 2l + 2w $, where $ l $ is the length and $ w $ is the width. Since we're given that the perimeter is 80 feet, we can write the equation as $ 80 = 2l + 2w $.

To find the maximum area, we need to maximize the area formula $ A = lw $. We can rewrite the perimeter equation to solve for one variable in terms of the other, such as $ l = 40 - w $ or $ w = 40 - l $, and substitute it into the area formula.

Let's use $ l = 40 - w $ and substitute it into the area formula:

$ A = l \times w = (40 - w) \times w $

Now, we have an area formula in terms of a single variable, $ w $. To find the maximum area, we can take the derivative of $ A $ with respect to $ w $, set it equal to zero, and solve for $ w $. Then, we can use that value to find the corresponding length $ l $.

After finding $ l $ and $ w $, substitute them back into the area formula to find the maximum area of the rectangle.