The curve is "bell-shaped"
http://www.wolframalpha.com/input/?i=plot+y+%3De%5E(-x%5E2)
Let the point of contact in quad I be P(x,y)
Did you mean "largest triangle" ??
If you want the "largest rectangle" then the tangent line is not needed.
Find the area of the largest rectangle cut from the first quadrant by a line tangent to the curve y=e^(-x^2)
3 answers
As Reiny says, a rectangle whose base goes from 0 to x has area
a = xy = xe^(-x^2)
da/dx = (1-2x^2) e^(-x^2)
so there is a max at x = 1/√2
That is assuming, of course that the rectangle has its base on the x-axis. It might be possible that a rotated rectangle could be bigger, but that gets too messy.
Oh, wow. Maybe that's what you meant. A rotated rectangle could have its top side tangent to the curve!
But that still gets messy, as issues of concavity and boundaries of the rectangle come into play.
a = xy = xe^(-x^2)
da/dx = (1-2x^2) e^(-x^2)
so there is a max at x = 1/√2
That is assuming, of course that the rectangle has its base on the x-axis. It might be possible that a rotated rectangle could be bigger, but that gets too messy.
Oh, wow. Maybe that's what you meant. A rotated rectangle could have its top side tangent to the curve!
But that still gets messy, as issues of concavity and boundaries of the rectangle come into play.
The answer is 2/e and I don't even have a clue on how to solve it.
0 answers