Question

I need help with optimizing!

What is the maximum possible area of a rectangle with a base that lies on the x-axis and two upper vertices on the graph y=e^((-x^2)/8)

Answer 1

Wow, what an interesting problem!!
Have you tried graphing this function?

Answer 2

Yes I have. I notice that as x gets very large and very small, the y value tends to 0 so I am wondering if a very small y value and a very large x value (giving me a very long skinny rectangle) will give me the largest area?

Answer 3

At first I thought to myself that an exponential function would not have the same y value for two different x-values, but then I noticed that your function is even, so this could happen.

Answer 4

Your best bet in optimization is to make use of the derivative. We need to write out an equation for the area of this rectangle, take the derivative of that, and then set the derivative equal to zero, to obtain the "critical values."

Answer 5

I've just graphed this function on my calculator and see that it is beautifully symmetrical. In fact, it looks like the "normal curve" that one encounters in statistics.

Answer 6

I've actually done that already by using the equation Area=wh, then finding the derivative and I found that the derivative is equal to 0 when x is positive or negative 2

Answer 7

And then I put that back into my area equation and found that the maximum possible area is 4/sqrt(e)

Answer 8

How would you know whether or not x=2 is correct as the value at which your rectangle has its max. area?

Answer 9

Well I guess I could check the second derivative

Answer 10

Here's the function I would write:
\[A=(2x)e ^{-x^2/2}\]

Answer 11

Same as yours?

Answer 12

yes that is what I started out with

Answer 13

oh wait no, it's A=(2x)e^(-x^2/8)

Answer 14

Cool! Were you to find the 2nd derivative now, and then substitute x=2 into it, your aim would be to determine wehther the 2nd derivative is positive or negative. What would it mean to you if the 2nd der. were neg.?

Answer 15

If the second derivative is negative then the graph is concave down and that would be giving me the maximum. I will calculate it now.

Answer 16

\[A=(2x)e^(-x^2/8)\] is what I get when I enter your expression into Equation Editor. I believe your intent here was the same as mine, exactly. 2x is the width of your rectangle, and e to the power (-x^2)/2 is the height.

Answer 17

"If the second derivative is negative then the graph is concave down and that would be giving me the maximum. I will calculate it now." Cool.

Answer 18

So will finding the maximum of this equation give me the maximum area? I know that with other questions I am supposed to find the endpoints to see if they end up giving me the maximum or the minimum, but I wasn't sure how to approach this question because there are no endpoints. The graph just gets infinitely closer to zero.

Answer 19

I did calculate the second derivative btw and it gave me a negative value when I input so it is a maximum. I am just wondering why wouldn't the x value be extremely large to produce a big area and how would I see if it is bigger than what I have already calculated?

Answer 20

The graph looks like this