The question is as follows. Given a house as shown in the figure below

enter image description here

$AB = 3a\,,\,BC = 10\,,\,CD = a\,,\,DE= a\,,\,EF=2a,$ and $FA=10-a$

Where a is some arbitrary constant such that $a \in [0,10]$ Now one could show that the area of the figure is given by the function

$$T(a) = 30a -2a^2$$

Whereas completing the square, or solving $T'(a)=0$, gives that the largest area of the house is $15^2/2$ for $a=15/2$.

Is there any other way of finding the largest area of said figure? (Hopefully using geometric reasoning.)

  • $\begingroup$ $\dfrac {225}{ 15}=15$ $\endgroup$ – Pedro Tamaroff Jul 5 '12 at 19:06
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    $\begingroup$ It's actually 225/2 not 225/15. $\endgroup$ – gt6989b Jul 5 '12 at 19:14

Here is an approach for the answer, that relies on a small (correct assumption) that would likely be possible to make more formal.

Step 1. We transform the problem into something more manageable. I would like to cut off the square on the top, together with the corresponding part of the rectangle, to get one rectangle with dimensions $2a \times (10-a)$ and that leaves another rectangle with dimensions $a \times 10$. Since we are into computing area, we can transform the second rectangle into an equivalent one with one side $2a$ and another side of $5$ (equivalent because halving one side and doubling another leaves the same area). Now connect the two rectangles together to get one rectangle of dimensions $2a \times (15-a)$.

Now we are asking how to geometrically prove that the maximum area of this rectangle happens at $a=15/2$.

Step 2. We know that if we need to maximize the area of a rectangle with a fixed perimeter of $30$, this results in finding the best $a$ such that rectangle with dimensions $a \times (15-a)$ contains the largest area. Geometrically, the optimal solution is a square with sides $a = 15-a = 15/2$.

This is exactly our problem, except we have one of the sides a linear scale larger (i.e. $2a$ instead of $a$). I would like to assume that such an extension (or any other simple multiplication of any of the sides by a constant) would not alter the correct solution.

Note: This assumption is correct, it is easy to verify using Calculus that since sucha transformation alters the objective function by a constant factor, it will have no effect on the location of the extrema, but proving this geometrically, as the author of the question requested, is non-trivial to me.

Hence the correct solution does indeed occur at $a=15/2$ and the maximum area will be $2a \times (15-a) = 15*15/2$ as desired.

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    $\begingroup$ Cut the rectangle of dimensions $2a\times(15-a)$ into two rectangles of dimensions $a\times(15-a)$ and your assumption is justified geometrically. :) $\endgroup$ – user856 Jul 5 '12 at 20:15
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    $\begingroup$ Alternately, note that applying an affine transformation to your rectangles (such as rescaling in the $x$-direction by a factor of 2) will not change the ratios of distinct rectangle areas; thus it will not change which rectangle maximizes the area. This is maybe a little less geometrically intuitive than Rahul's solution, but it quickly gives the fully general result with no calculus involved. $\endgroup$ – Micah Jul 5 '12 at 21:26

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