Expert: Paul Klarreich - 10/20/2008

Question
Problem: Consider a rectangle with its base on the x-axis and two vertices on the curve f(x) = e^(-(x^2). Show that the rectangle has the largest possible area when the two vertices are the points of inflection of the graph of f.

We have already covered derivative, optimization, as well as L'Hopital and have just finished integrals. Optimization has always been my weak point.

Answer
Questioner:   Zahra
Category:  Calculus
Private:  No
 
Subject:  BC Calculus
Question:  Problem: Consider a rectangle with its base on the x-axis and two vertices on the curve f(x) = e^(-(x^2). Show that the rectangle has the largest possible area when the two vertices are the points of inflection of the graph of f.

We have already covered derivative, optimization, as well as L'Hopital and have just finished integrals. Optimization has always been my weak point.
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Hi, Zahra,

Is this your first attempt at Max-min problems?  If so, the scheme is something like this:

1. Identify the variable(s) in the problem -- the things that can be changed.  Give them names.  A diagram might be good.

2. Write the thing to be Max-ed(or min-ed) in terms of the variable(s).

3. If there are more than one, determine a relationship between the variables (called a 'CONSTRAINT') that will eliminate all but one.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is probably the hardest part.

4. Now differentiate, set the derivative = 0, and solve.  Be sure to check for endpoint max-mins.

AND, Please check the archives for other M-M examples.  There are a lot of them.  Click BROWSE PAST ANSWERS.

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Let  x be the right-hand corner (see diagram)

>> sorry. I forgot to attach it, and the site won't let me do it.  If you resubmit it as a followup, I can do it.

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Then the base is 2x.
And the height is  y = exp(-x^2)
And the area A = 2xy = x exp(-x^2)

Differentiate:

dA/dx = (x)(-2x exp(-x^2)) + (1)(exp(-x^2))

dA/dx = - 2x^2 exp(-x^2)) + exp(-x^2)

Set that = 0 and solve:

- 2x^2 exp(-x^2) + exp(-x^2) = 0

exp(-x^2)( - 2x^2 + 1) = 0

The exp(..) is never zero, so set:

-2x^2 + 1 = 0

x^2 = 1/2

x = +- sqrt(2)/2

So that is where A is maxed.
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Now as to your proof.  Inflection points occur at  y'' = 0.

y = exp(-2x^2)

y' = -2x exp(-2x^2) = -2 times (x exp(-2x^2)) = -2 times Area.

y'' = -2 times the derivative of area we got above.

Cute.

You can handle it from here.