# Optimization and Tangent Line Question

I'm stuck on a question involving maximizing the area of a triangle:

What is the area of the largest triangle that can be formed in the first quadrant by the x-axis, the y-axis, and a tangent line to the graph $y=e^{-x}$ ?

I made a diagram for myself in order to better understand the question. Following is what I have come up with so far.

So I am looking to maximize the area of the triangle that is shaded in light blue. The area of the triangle is of course $A=0.5 bh$. The height will be the value of the tangent line where $x=0$, and the base of the triangle will be the length from the y-axis to the point where the tangent line intercepts the x-axis. I have found the tangent line for the point $a$:

$$y=e^{-a}-e^{-a}(x-a)$$

So the height of the triangle should be $y=e^{-a}-e^{-a}(0-a)$. I am unsure how to proceed and find the length from the y-axis to the point where the tangent line intercepts the x-axis (the inverse of the tangent line?). I believe that after I find the equation for the area of the triangle I will need to the find the maximum value.

Thank you.

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How did you make that picture? – yiyi Oct 26 '12 at 1:06

You found the height of the triangle correctly. It simplifies to $(a+1)e^{-a}$.
To find the base of the triangle, put $y=0$ in the equation of the tangent line, and solve for $x$. There is very nice cancellation, and we get $x=a+1$. Thus the area $A(a)$, as a function of $a$, is given by $$A(a)=\frac{1}{2}(a+1)^2 e^{-a}.$$ There should now be no difficulty in maximizing the area. (I would prefer to maximize $2A(a)$, no fractions.) Calculate the derivative with respect to the variable $a$ as usual, and argue that one of the two critical points gives maximum area. We have to be a little careful about the geometry: if $a< -1$, there is no triangle in the first quadrant. The formula for $A(a)$ cheerfully gives you an "area" even when $a<-1$.
There is a typo in the picture. The equation of the tangent line at the point $(a,e^{-a})$ should be $y=e^{-a}+f'(a)(x-a)$.