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I have a calculus final coming up and was going over a practice exam my professor gave me and I came across a problem I was struggling with. I would post a picture but I am having trouble posting a picture So the problem reads.

The rectangle shown here has one side on the positive $x$-axis, one side on the positive $y$-axis, and its upper right hand vertex on the curve $y=e^{-x^2}$. What dimensions give the rectangle its largest area?

Keep In mind I already have the answer that’s not what I’m looking for, I’m searching for the process to arrive at the answer and the necessary work. BTW The answer is $x=1/\sqrt{2}$ and $y=e^{-1/2}$. Thank you for the help.

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  • $\begingroup$ 1) Draw a picture 2) Set up some parameters ($x, y$ here of course) 3) Identify the constraint ($y=e^{-x^2}$) here 4) Identify the quantity to optimize ($xy$ here) 5) Make the latter a function of one variable thanks to the constraint 6) Study that function to find the extremum $\endgroup$ – Julien Dec 13 '13 at 4:29
  • $\begingroup$ Thank you for your reply it was very helpful, however I have one more question would finding the constraint on an equation like this entail finding the derivative of the curve's equation set to equal 0. I couldn’t figure out if that would give me the vertex of the curve,I.E. the upper right hand corner of the box. $\endgroup$ – user3097903 Dec 13 '13 at 4:42
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Your rectangle is defined by the four points (0,0), (x,0), (x,y) and (0,y). The area of the rectangle i defined by A = x y but the right corner must be along the curve y = Exp[-x^2]. So, the area is given by A = x Exp[-x^2] and this must be maximized. Then, as usual, we go to the derivative which is
dA/dx = Exp[-x^2] (1 - 2 x^2)
At the extremum, the derivative must be zero and then this happens for x = 1 / Sqrt[2] and x = -1 / Sqrt[2]. Since the problem is symetric, we shall only consider the positive value. For this value of x, then y = Exp[-x^2] gives Exp[-1/2] and the maximum area is 1 / Sqrt[2 e].
We finally need to confirm that this is a maximum; so, as usual, we go to the second derivative which is
d2A/dx2 = x Exp[-x^2] (2 x^2 - 3)
For x = 1 / Sqrt[2], the second derivative is then negative and this confirms that we obtained a maximum value.

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  • $\begingroup$ Thank you very much that was extremely helpful. $\endgroup$ – user3097903 Dec 13 '13 at 4:45
  • $\begingroup$ @user3097903. I am glad to have been able to help. I hope you noticed that the problem is simple and almost involves natural language plus very few mathematics. Cheers. $\endgroup$ – Claude Leibovici Dec 13 '13 at 4:48

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