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I am stuck on this optimization problem:

I am supposed to find the largest rectangle that can be fit into the right triangle. I think I am having trouble with setting up the constraints. I am able to optimize cylinders in spheres but I cannot seem to figure this one out. So far my constraints are as follows: $$\begin{align}A_{rect}&=(11-w)(60-h)\\A_{tri}&=\dfrac{1}{2}wh\end{align}$$

I looked over this again and it seems that my equation for the area of a rectangle isn't correct. I cannot figure out a way write the constraint for the rectangle. Is there another method to solving this problem?

grg
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Kot
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  • It seems to me that $A_{rect} = wh$. Your second constraint might then be $\sqrt{(11-w)^{2} + h^{2}} + \sqrt{w^{2} + (60-h)^{2}} = 61$. – Alex Wertheim May 08 '13 at 03:47
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    Or by similarity, the simpler $11h = 60 (11-w)$ – ronno May 08 '13 at 03:52
  • @AWertheim I thought that the constraint equation had to be for the area of the triangle. You could have a constraint for the hypotenuse? – Kot May 08 '13 at 03:53
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    You aren't being asked to optimize the triangle anyway, so you don't need an area formula for it. As AWertheim says, you want to maximize $A_{rect} = wh$ , just as you've labeled it. However, it would make life easier to express the height $h$ of the rectangle in terms of the equation for a line represented by the hypotenuse of that right triangle. Pretend the left lower corner is the origin: what is the slope of the hypotenuse? What would be an equation for the line that the hypotenuse lies along? You would have $ h = f(11 - w)$ , which you can use to replace $w$ in your area function. – colormegone May 08 '13 at 03:54
  • Good thinking, @ronno, was a bit lazy in my haste ;) – Alex Wertheim May 08 '13 at 03:56
  • I see, if by similarity. Would $60w=11(60-h)$ be equal to $11h=60(11-w)$? – Kot May 08 '13 at 04:01
  • Yes, you could use either of those equations, depending upon which variable you wish to "eliminate" in the rectangle's area function. (If you multiply out the terms in your two similarity equations, you'll see that they are just rearrangements of one another.) – colormegone May 08 '13 at 06:29
  • It's not clear to me from the comments whether the question has been answered. If so, please consider writing up an answer and accepting it so that the question doesn't remain marked as unanswered. If it has only partially been answered, please update the question to make it clear what part of it remains open. – joriki May 08 '13 at 07:42

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