I was studying for some quizzes when I encountered this problem. It goes like this:
Find the area of the largest triangle cut from the first quadrant by a line tangent to the curve $y = e^{-x}$
My work:
This is how I imagined the problem:
The area of the triangle as described in the picture above is $A = \frac{1}{2} x y.$ Since $y = e^{-x},$ then the area would be
$A = \frac{1}{2} x e^{-x}$
We need to maximize the area, so we get the derivative of the area $A.$
The derivative of $A = \frac{1}{2} x e^{-x}$ is $\frac{dA}{dx} = \frac{1}{2} (-xe^{-x} + e^{-x})$ or:
$$\frac{dA}{dx} = e^{-x} \left( \frac{-x}{2} + \frac{1}{2} \right)$$
To get the maximum area of the triangle, we set $\frac{dA}{dx}$ to zero.
$$\frac{dA}{dx} = e^{-x} \left( \frac{-x}{2} + \frac{1}{2} \right)$$
$$0 = e^{-x} \left( \frac{-x}{2} + \frac{1}{2} \right)$$
So we got: $e^-x = 0$ and $ \frac{-x}{2} + \frac{1}{2} = 0 $
Getting the value of $x$ in the expression $e^{-x} = 0,$ $x$ is undefined.
Getting the value of $x$ in the expression $\frac{-x}{2} + \frac{1}{2} = 0,$ $x =1$.
To verify that $x = 1$ is the value that gives the maximum area, we do a First Derivative Test. Making the $x =1 $ as the critical/ base point, we try $x = 0.32$ and $x = 99.$
If $x = 0.32,$ the the value of the derivative is $$\frac{dA}{dx} = e^{-(0.32)} \left( \frac{-(0.32)}{2} + \frac{1}{2} \right) = + 0.2469$$
If $x = 99,$ the the value of the derivative is $$\frac{dA}{dx} = e^{-(99)} \left( \frac{-(99)}{2} + \frac{1}{2} \right) = -4.955 \times 10^{42}$$
If the sign of derivative changes from (+,0,-), then the critical point is the maximum point, which is $x = 1.$
At this point, I got the slope of 1, which comes from $x = 1.$ I don't know what to do next. How do you get the area of the
largest triangle that can be cut from the first quadrant by a line tangent to the curve $y = e^x$ ?
Best Answer
Presumably, you want the area of the triangle in the first quadrant bounded by the tangent line and the axes.
Based on that assumption, if $(x,y) = (t,e^{-t})$ is the point of tangency, the area is not $\frac{1}{2}xy$.
Instead, leaving $t$ unknown, find the equation of the tangent line in terms of $t$, then find the $x$ and $y$ intercepts, also in terms of $t$. Once you have the intercepts, you can express the area of the triangle as a function of $t$. That's the function you want to maximize.
Can you finish it?