Tangent of $y=e^{-x},x\ge-1$ crosses the $x$ and $y$ axis in points $A,B$. What is the maximum possible area of the $OAB$ triangle ($O(0,0)$)?

Question

Tangent of $y=e^{-x}, x\ge-1$ crosses the $x$ and $y$ axis in points $A,B$. What is the maximum possible area of the $OAB$ triangle?

I figure it has to do with derivatives, but I'm not sure how to express the surface area of the triangle well. The tangent formula would be $y=-\frac{x}{e^{x0}}+n$ but I'm really just stuck. Any help is much appreciated.

Answer 1

Tangent at $x_0$: $$y=e^{-x_0}-e^{-x_0}(x-x_0).$$ Area: $$S=\frac{AB}{2}=\frac{(1+x_0)^2}{2e^{x_0}}.$$ $$S'=0 \Rightarrow x_0=\pm 1.$$ Tangent can not be drawn at $x_0=-1.$ $$S''(1)<0 \Rightarrow S(1)=\frac{2}{e} (max).$$

Answer 2

You can find the equation of the line, tangent to the curve, at any point $x_0<-1$. Then, find the intersection of the line with $x$ and $y$ axis, simply by putting $y$ and $x$ equal to zero, in the line equation. Call the solutions $a$ and $b$. The area of the triangle is $f(x_0)=\frac{a\times b}{2}$, which is a function of $x_0$. Now, you need to optimize $f(x_0)$.

The condition $x<-1$ makes sure that $a$ and $b$ would not change sign, so you do not need absolute value in the area formula.