Posted by Alex on Monday, January 3, 2011 at 5:59pm.
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)
Can you show me the steps you skipped?
Thanks
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
Related Questions
Calculus - Find the dimensions of teh rectangle with the most area that will ...
Calculus - Hello, could someone please help me with this problem? I'm a ...
Calculus - Find the dimensions of the rectangle of largest area that has its ...
Calculus - Find the maximum area of a rectangle that lies above the x axis and ...
calculus - A rectangle with its base on the x-axis is to be inscribed under the ...
calculus - The graph of f'(x) is shown for 0=< x =<10. The ...
calculus - A rectangle with its base on the x-axis is to be inscribed under the ...
Calculus - Please help? Let F(a) be the area between the x-axis and the graph ...
Calculus - Please help? Let F(a) be the area between the x-axis and the graph of...
Calculus - Let F(a) be the area between the x-axis and the graph of y=x^2cos(x/4...
For Further Reading
