# Thread: Help with max and min.

1. ## Help with max and min.

A right triangle has one leg on the x-axis.The vertex at the right end of that leg is at the point (3,0).The other vertex touches the graph of y=e^x.The entire triangle is to lie in the first quadrant.Find the max area of this triangle.

Thanks.

2. Originally Posted by jupiter92
A right triangle has one leg on the x-axis.The vertex at the right end of that leg is at the point (3,0).The other vertex touches the graph of y=e^x.The entire triangle is to lie in the first quadrant.Find the max area of this triangle.

Thanks.
From what you have described, it would appear that your triangle has vertices:

$(3, 0), (x, e^x), (x, 0)$ and I assume that $x < 3$.

Therefore its base is $3 - x$ and its height is $e^x$.

So its area is given by the formula:

$A = \frac{1}{2}(3 - x)e^x$.

Since you want the maximum area:

$\frac{dA}{dx} = \frac{1}{2}\left[(3 - x)e^x - e^x\right]$

$= \frac{1}{2}e^x(3 - x - 1)$

$= \frac{1}{2}e^x(2 - x)$.

Setting the derivative equal to $0$ and solving for $x$:

$\frac{1}{2}e^x(2 - x) = 0$

$e^x = 0$ or $2 - x = 0$.

But since $e^x > 0$ for all $x$, that means that only $2 - x = 0$ is valid.

Therefore $x = 2$.

To check that this IS a maximum, find the second derivative and check that it is negative at $x = 2$.

$\frac{d^2A}{dx^2} = \frac{1}{2}\left[e^x(2 - x) - e^x\right]$

$= \frac{1}{2}e^x(2 - x - 1)$

$= \frac{1}{2}e^x(1 - x)$

At $x = 2$

$\frac{d^2A}{dx^2} = \frac{1}{2}e^2(1 - 2)$

$= -\frac{1}{2}e^2$.

This function is negative at $x = 2$, so $x = 2$ does indeed maximise the area.

So at $x = 2$:

$A = \frac{1}{2}(3 - 2)e^2$

$= \frac{1}{2}e^2$ square units.

3. thanks so much.