Question

Take a point P(a,e^-a) (a> -1) on the curve C: y=e^-x, Let S(a) be the area of the triangle surrounded by the tangent line to C at P, the x-axis and the y-axis.
*Find the function S(a).
*Find the maximum of S(a).

Answer 1

First find the equation of the tangent line through \(P\).
\[y(x)=e^{-x}\implies y'(x)=-e^{-x}\implies y'(a)=-e^{-a}\]So the equation of the tangent line is
\[y-e^{-a}=-e^{-a}(x-a)\implies y=-e^{-a}x+(a+1)e^{-a}\]Find the \(x\) and \(y\) intercepts of this line. These points will tell you what the base and height of the triangle will be.
\[x=0\implies y=(a+1)e^{-a}\\
y=0\implies x=a+1\]The area of the triangle is then
\[S(a)=\frac{1}{2}(a+1)^2e^{-a}\]Find the critical points of \(S\):
\[S'(a)=\frac{1}{2}\left(2(a+1)e^{-a}-(a+1)^2e^{-a}\right)=\frac{1}{2}(1-a^2)e^{-a}\]Which is \(0\) at \(a=\pm 1\). Discard \(a=-1\), since we're focusing on \(a>-1\). Check to make sure a maximum indeed occurs at this critical point by checking the sign of the derivative to either side. Once you confirm that, find the actual value of \(S\) at \(a=1\).

Answer 2

the intersection looks like this

Answer 3

which is why the x and y intercepts are the legs of the triangle,
S(A)=1/2xy

Answer 4

@HolsterEmission Thank you for your answer, it helped me a lot! :D