Posted by Alex on .
Find the dimensions of the rectangle with the most area that will fit above the xaxis and below the graph of y=e^x^2.

Calculus 
Reiny,
The curve looks sort of bellshaped.
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,
Can you show me the steps you skipped?
Thanks 
Calculus 
Reiny,
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