Find the maximum area of the rectangle confined between g(x) = $e^{\frac{1}{2}x}$, f(x) = $e^{-x}$, and $x$- axis. (photo and figure included)

Question

Find the maximum area of the rectangle confined between $g(x) = e^{\frac{1 }{2}x}, f(x) = e^{-x}$, and the $x-$ axis. Here I attached a picture with the figure illustrated and how I attempted to solve the problem. First I assigned different x's to each function: $(x_2, e^{\frac{1}{2}x_2})$ and $(x_1, e^{-x_1})$. Then I formed an equation for the area of the rectangle: $(x_1 + x_2)(e^{\frac{1}{2}x})$. Next I tried solving for $e^{\frac{1}{2}x_2}$ = $e^{-x_1}$ to be able to represent one $x$ with another $x$ so I can substitute it in the equation. Next, I found the derivative of the area equation and equaled it to zero so I get the x-coordinate of the maximum point but I ended up getting the x-coordinate of where the two functions intersect. Please see image to better understand the problem and see the full solving process.

Answer 1

Solution

Let $(x_1,e^{-x_1})(x_1>0)$ lie on $y=e^{-x}$,and $(x_2,e^{x_2/2})(x_2<0)$ lie on $y=e^{x/2}$.

Notice that we need a rectangle. Hence, $$ e^{-x_1}=e^{x_2/2}.$$ It's clear that $$-x_1=\frac{x_2}{2},$$namely, $$x_2=-2x_1.$$

Thus, for the area $A$, we have $$A=(x_1-x_2)e^{-x_1}=3x_1e^{-x_1},~~~x_1>0.$$

Denote $f(x)=3xe^{-x}(x>0).$ Then $f'(x)=\dfrac{3(1-x)}{e^x}$. We may know readily that $f(x)$ reaches its maximum value at $x=1$. It follows that $$\max (A)=\max_{x>0}f(x)=f(1)=\frac{3}{e}.$$