# Calculus

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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 - ,

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 - ,

Can you show me the steps you skipped?
Thanks

• Calculus - ,

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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