The problem says that we have to find the maximum area of a rectangle that circumscribes another rectangle of sides $a$ and $b$.
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What choices exist for the circumscribing rectangle? – mvw Mar 23 '17 at 11:58
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The circumscribing rectangle must have one vertex od the original rectangle in each side of it – Josemi Mar 23 '17 at 12:12
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A solution based on trigonometry:
Let $\alpha$ be the (smallest) angle formed by a side of the original rectangle and a side of the circumscribing rectangle. Then the area $A$ of the circumscribing rectangle is equal to the area $ab$ of the original rectangle plus the area of four right triangles, $a\sin\alpha\cdot a\cos\alpha + b\sin\alpha\cdot b\cos\alpha$. So $$ A = ab + a^2\sin\alpha\cos\alpha + b^2\sin\alpha\cos\alpha = ab + (a^2+b^2)\sin\alpha\cos\alpha $$ $$ = ab + {1\over2}(a^2+b^2)\sin2\alpha. \quad\mbox{(We used: } \ \sin2\alpha=2\sin\alpha\cos\alpha.) $$ The area $A$ attains its maximum when $\sin2\alpha=1$, i.e. when $\alpha=45^\circ$, and we have $$ A_\max = ab + {1\over2}(a^2+b^2) = {(a+b)^2\over2}. $$
(It is easy to check that the circumscribing rectangle with area $A_\max$ is a square with side ${a+b\over\sqrt2}$.)
Alex
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