Maximize Rectangle area with y=e^-x/6

Question

A rectangle has one corner at the origin and the other on the curve . Find the dimensions that maximize the rectangles area. Assume  is positive

Width x=

Height y=

Answer 1

The area of the rectangle is A=xe^-x/6

Then, dA/dx= e^-x/6 - 1/6xe^-x/6 = (1 - 1/6x)e^-x/6

This equals 0 when (1 - 1/6x) = 0, or x = 6

d²A/dx² = (-1/3 + 1/36x)e^-x/6

As this is less than 0 for x < 12, the Area is a maximum at x = 6

The maximum area is 6e^-1

Answer 2

you are trying to maximize the area of a rectangle, we can call the width x and the height y so:

area=xy

using the equation above we know:

y=e^-x/6

so lets plug y into the area equation and then take the derivative:

area=xe^-x/6, area'=e^-x/6-x/6(e^-x/6)

now we set this equation equal to 0 and solve for x

0=e^-x/6-x/6(e^-x/6), e^-x/6=x/6(e^-x/6) (divide out the e^-x/6), 1=x/6, x=6

Now we plug in 6 for x in the y=e^-x/6 equation to solve for y:

y=e^-6/6, y=e^-1=1/e

and there you have your dimensions