Maximum & Minimum: Greatest Possible Area

Question

Let  be a point on the graph of  with . Let  be a rectangle with one side on the -axis and two vertices on the graph, as shown in the picture next. Find the rectangle with the greatest possible area. Enter the exact value of the area of this rectangle. Greatest area:  ___unit²

Answer 1

Write the equation of the area of the rectangle (this is the function you are trying to maximize).

A = height•width

The height is the y-value which is "-x² + 8".

The width is 2x.

So the area is (-x² + 8)(2x)

A(x) = -2x³ + 16x

To find the extrema, take the derivative and set it equal to zero:

A'(x) = -6x² + 16

-6x² + 16 = 0

6x² = 16

x = ±√(16/6) = ±4/√6 = ±4√6/6 = ±2√6/3

To find out if you have a minimum or a maximum, you CAN do the 1st derivative test however a negative cubic comes from the upper left and goes toward the bottom right of the graph with a minimum at the left and a maximum at the right therefore, -2√6/3 is a minimum and 2√6/3 is a maximum.

So the area is found by plugging in the x-value into the area function:

A(x) = -2x³ + 16x

A(2√6/3) = -2(2√6/3)³ + 16(2√6/3)

A(2√6/3) = -2(8√216)/27 + 32√6/3 = -96√6/27 + 288√6/27 = 192√6/27 = 64√6/9 square units