Calculus Optimization Problem

Question

Find the maximum area of a rectangle in the first quadrant under the function y=3e^-x/2.

Answer 1

Express A in terms of x then maximize. Length of the rectangle along the x-axis is x. The height of the rectangle at x is y=3e^-x/2.

A = x * y = 3xe^-x/2

A(x) = 3xe^-x/2

Take the first derivative and set it equal to zero to find critical points:

A'(x) = 3e^-x/2(1-x/2) = 0

3e^-x/2 can never equal 0, so we solve (1-x/2) = 0

x = 2

Use the Second Derivative Test to see if x = 2 is a local maximum/minimum:

A''(x) = 3e^−x/2(x/4−1)

Since A''(2) < 0, the critical point at x = 2 is a local maximum.

Substitute x = 2 back into the A function:

A(2) = 3⋅2e^−2/2 = 6/e

The maximum area of the rectangle is 6/e.

Hope this was helpful.