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.

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- Mar 6th 2010, 05:57 PM #1

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## 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.

- Mar 6th 2010, 08:19 PM #2
From what you have described, it would appear that your triangle has vertices:

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

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

So its area is given by the formula:

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

Since you want the maximum area:

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

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

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

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

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

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

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

Therefore $\displaystyle x = 2$.

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

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

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

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

At $\displaystyle x = 2$

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

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

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

So at $\displaystyle x = 2$:

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

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

- Mar 13th 2010, 12:17 PM #3