Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
$\dfrac {225}{ 15}=15$ –  Pedro Tamaroff Jul 5 '12 at 19:06
2  
It's actually 225/2 not 225/15. –  gt6989b Jul 5 '12 at 19:14
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
2  
Cut the rectangle of dimensions $2a\times(15-a)$ into two rectangles of dimensions $a\times(15-a)$ and your assumption is justified geometrically. :) –  Rahul Jul 5 '12 at 20:15
1  
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. –  Micah Jul 5 '12 at 21:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.