A problem for all Calculus whizzes

[MHENK]

This, like stated above, is a problem for calculus whizzes. I am STUCK..

Q What is the greatest possible area of a triangle formed in the first quadrant by the x-axis, the y-axis, and a line tangent to the curve y=e^-x?

Any help would be appreciated. If it's not too much to ask, could i possibly see the work done when figuring out this problem. I don't really just want the answer, so if you could even point me in the direction, that'd be okay too..

-MHENK

[Guv]

I have not been meticulous about the following, so do not assume that I have it correct. When not getting paid with money or sexual favors, I do not guarantee my work.


Function(x) = e^-x

Derivative(x) = -e^-x

Assume a point on the curve (p, e^-p)

Equation of tangent line is y = m*x + y0
where m is the derivative at x = p and y0 is the y-intercept (y when x = 0).

Substitute for m, resulting in the following.

Equation of tangent line is y = -x*e^-p + y0

(P, e^-p) is on the curve as well as on the tangent line. Therefore the following.

e^-p = -p*e^-p + y0, which can be rearranged to provide the following.

y0 = e^-p + p*e^-p, which is y-intercept.

y0 = (1 + p)*e^-p Substituting in tangent line equation yields the following.

y = -x*e^-p + (1+p)*e^-p

When y = 0, 0 = -x*e^-p + (1 + p)*e^-p

x*e^-p = (1 + p)*e^-p

x = 1 + p is the x-intercept.

TriangleArea = x-intercept*y-intercept / 2

TriangleArea = [ (1 + p)^2 ]*e^-p / 2

Now find value of p for which the above is a maximum. This requires taking derivative of TriangleArea with respect to p and finding a zero value. I am not sure of the implications of there being no value of p for which the derivative is zero.

At this point, I have done a lot of thinking for no compensation. When I do not get paid in money or sexual favors, I get lazy. Can you do the rest from here?

[wossname]

I don't even guarantee my work with sexual favors and money!