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Find the maximum & minimum area of an outer rotated rectangle when the inner rectangle has the side lengths $a$ and $b$.

Here's an image:

two rectangles

I have already tried to relate the side of outer triangle with inner triangle by denoting an angle between them, but I couldn't get all the sides.

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  • $\begingroup$ inside or outside ??? $\endgroup$ – Yves Daoust May 27 '15 at 12:18
  • $\begingroup$ @YvesDaoust Outside, the inside is constant. $\endgroup$ – Mario May 27 '15 at 12:18
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    $\begingroup$ The minimum outside rectangle is the rectangle itself !!! $\endgroup$ – Yves Daoust May 27 '15 at 12:21
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    $\begingroup$ @Mario It is still unclear what we are allowed to do with the outer rectangle, what configurations are feasible? Once we understand this we can help picking one with minimal area out of them. $\endgroup$ – mvw May 27 '15 at 12:27
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    $\begingroup$ @mvw: actually, guessing what the OP means, the angle is fixed and there is no minimum to find, just the area of the bounding box. $\endgroup$ – Yves Daoust May 27 '15 at 12:56
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As others noted (when the question only asked for the minimum), the minimum area obviously occurs when the outer rectangle is identical with the inner. As for the maximum, note that the four right triangles surrounding the inner rectangle are all similar, so it suffices to find the largest right triangle with hypotenuse of prescribed length ($a$ for the small ones, $b$ for the large ones, assuming $a\lt b$). This is easily seen to occur for the isosceles right triangle, with sides of length $a/\sqrt2$ and $b/\sqrt2$, respectively. This means the area-maximizing rectangle is a square with sides of length $(a+b)/\sqrt2$, hence area $(a+b)^2/2$.

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Suppose the first rectangle (call it $R$) has vertices $(0,0),\, (a,0),\, (a,b),\, (0,b).$ Then, if I have understood your question correctly, we want to find the minimum area of another rectangle, $R'$ say, containing $R$. We assume that the vertices of $R$ touch the edges of $R'$. We determine the area of $R'$ given that it is at an angle of $0\le\phi<\frac\pi2$ to $R$.

When rotated by an angle $\phi$, the vertices of $R$ become $$\binom 00,\,\binom{a\cos\phi}{a\sin\phi},\,\binom{a\cos\phi-b\sin\phi}{a\sin\phi+b\cos\phi},\,\binom{-b\sin\phi}{b\cos\phi}$$

We draw $R'$ with sides parallel to the axes so it has width $$b\sin\phi+a\cos\phi,$$ and height $$a\sin\phi+b\cos\phi.$$

Therefore it has area $$A=(a\sin\phi+b\cos\phi)(a\cos\phi+b\sin\phi) = ab + (a^2+b^2)\sin\phi\cos\phi$$ but we note that $\sin\phi\cos\phi=\frac12\sin2\phi$ so the area is minimised when $\phi=0,\frac\pi2$ and takes a value of $ab$.

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    $\begingroup$ Do we need all this to see that the minimum rectangle is the original rectangle ? (Not counting the fact that the minimum of your $A$ is $ab-a^2-b^2$.) $\endgroup$ – Yves Daoust May 27 '15 at 12:29
  • $\begingroup$ Not really but I'm expecting that something has been missed out of the problem that will probably put some extra constraints on $\phi$ $\endgroup$ – Dan Robertson May 27 '15 at 12:34
  • $\begingroup$ You should deal with that using absolute values in the evaluation of the sides, there can be no negative term. $\endgroup$ – Yves Daoust May 27 '15 at 12:36
  • $\begingroup$ I'm avoiding that by keeping $0\le\phi<\frac\pi2$ as the situation is the essentially periodic in $\phi$ $\endgroup$ – Dan Robertson May 27 '15 at 12:42
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By elementary trigonometry, the base of the outer rectangle is

$$a\lvert\cos(\theta)\rvert+b\lvert\sin(\theta)\rvert,$$ for any angle. The height can be obtained by rotating by an extra quarter turn

$$a\lvert\cos\left(\theta+\frac\pi2\right)\rvert+b\lvert\sin\left(\theta+\frac\pi2\right)\rvert=a\lvert\sin(\theta)\rvert+b\lvert\cos(\theta)\rvert.$$

The area equals $$ab+(a^2+b^2)\lvert\cos(\theta)\sin(\theta)\rvert$$ after simplification, going from $ab$ (obviously) to $\dfrac12(a+b)^2$.

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