Posted by **Alex** on Monday, January 3, 2011 at 5:59pm.

Find the dimensions of the rectangle with the most area that will fit above the x-axis and below the graph of y=e^-x^2.

- Calculus -
**Reiny**, Monday, January 3, 2011 at 6:10pm
The curve looks sort of bell-shaped.

Let the point of contact in quadrant I be (x,y)

then the base of the rectangle is 2x and its height is y

A = 2xy

= 2x(e^(-x^2))

d(A)/dx = (2x)(-2x)(e^(-x^2)) + 2(e^(-x^2))

= 0 for a max/min of A

skipping some steps

e^(-x^2)) ( -2x + 1) = 0

x = 1/2

then y = e^(-1/4) = 1/e^(1/4)

so the rectangle has a length of 2x or 1

and a height of 1/e^(1/4)

- Calculus -
**Alex**, Monday, January 3, 2011 at 6:41pm
Can you show me the steps you skipped?

Thanks

- Calculus -
**Reiny**, Monday, January 3, 2011 at 7:37pm
from

(2x)(-2x)(e^(-x^2)) + 2(e^(-x^2)) = 0

-2e^(-x^2) ( -2x^2 + 1) = 0

**ahhh, just noticed an error from earlier, (that's what I get by skipping steps)**

-2e^(-x^2) = 0 ----> no solution

or

-2x^2 + 1 = 0

x^2 = 1/2

x = 1/√2

then y = e^(-1/2) = 1/√e

so the rectangle has a length of 2x or 2/√2

and a height of 1/√e

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