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Problem of the Week #53 - April 1st, 2013

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Chris L T521

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Jan 26, 2012
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I was on vacation last week and forgot to mention that we hit a milestone with regard to the Problem of the Week on MHB - we're 52 weeks (1 year) strong and haven't missed a single week since we started the POTWs (Jameson luckily mentioned all of this in his post for the 52nd Secondary School POTW). We are now entering the 53rd week of the University POTW on MHB, and if it weren't for you guys participating, I'm not sure we would gotten this far. With that said, here's to another year of POTWs and many more in the future! At this time, I would also like to remind you that if you have a question you'd like to submit for the POTW (any level), you can do so by clicking on the POTW tab at the top of the forum page and fill out the appropriate form. Thanks a bunch guys for making this a successul part of the MHB experience!


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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: A rectangular box is to be inscribed in the tetrahedron whose faces are the coordinate planes and the plane $x/a+y/b+z/c=1$ (where $a$, $b$, and $c$ are positive constants). One corner of the box touches the plane, the opposite corner is at the origin, and the faces of the box are parallel to the coordinate planes. Find the dimensions of the largest such box.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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This week's problem was correctly answered by Opalg and Sudharaka.

Here's Opalg's solution (which utilizes Lagrange Multipliers):
If the vertices of the box that lie on the coordinate axes are at the points $(x,0,0),\ (0,y,0),\ (0,0,z)$, then the volume of the box is $xyz$. The vertex opposite the origin is at the point $(x,y,z)$, and that must lie on the given plane. So we want to maximise $xyz$ subject to the constraint $\dfrac xa + \dfrac yb + \dfrac zc = 1$.

Using the method of Lagrange multipliers, let $f(x,y,z,\lambda) = xyz - \lambda\bigl(\frac xa + \frac yb + \frac zc - 1\bigr)$. Put the partial derivatives equal to $0$: $$\frac{\partial f}{\partial x} = yz - \frac\lambda a = 0 , \qquad \frac{\partial f}{\partial y} = xz - \frac\lambda b = 0, \qquad \frac{\partial f}{\partial z} = xy - \frac\lambda c = 0.$$ Therefore $xyz = \lambda\dfrac xa = \lambda\dfrac yb = \lambda\dfrac zc$, from which it follows that $\dfrac xa = \dfrac yb = \dfrac zc$ (unless $\lambda=0$, which would imply $xyz=0$, in which case we get a minimum rather than a maximum for the volume). The condition $\dfrac xa + \dfrac yb + \dfrac zc = 1$ then implies that $\dfrac xa = \dfrac yb = \dfrac zc = \dfrac13.$

So the only extremal point occurs when $(x,y,z) = \frac13(a,b,c)$, and the volume is then $\frac1{27}abc$. To check that this is a maximum, notice that the boundary of the region of constraint lies entirely on the coordinate axes, and the volume of the box is $0$ if any one of $x,\ y,\ z$ is $0$. So the extremal point must necessarily be a maximum.

Here's Sudharaka's solution (without Lagrange Multipliers):
Let \((x,\,y,\,z)\) be the coordinates of the corner that touches the plane. Then the volume of the box is,

\[V=xyz\]

Also we have,

\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]

By the above two equations we get,

\[V=xy-\frac{x^2 y}{a}-\frac{xy^2}{b}\]

Now using the Second Partial Derivative Test we obtain that, \(V\) has a maximum at \(x=\dfrac{a}{3},\,y=\dfrac{b}{3},\,z=\dfrac{c}{3}\). Therefore the dimensions of the largest box is, \(\displaystyle\left(\frac{a}{3},\,\frac{b}{3},\, \frac{c}{3}\right)\) where \(a,\,b,\,c>0\).
 
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