# Math Help - MAX volume given array of constraints

1. ## Re: MAX volume given array of constraints

$A=\frac{1}{2}zy$ where $z=ky, k\geq0$
Thus,
$A(k, y)=\frac{1}{2}ky^2$
and
$A_1=\frac{1}{2}y^2$

$A_2=ky$
Not entirely sure how you got k=1 from these partials. Is that not how you determine the max of the area?

2. ## Re: MAX volume given array of constraints

$x+y(k+1)+y\sqrt{k^2+1} \le 2$
So, you want to maximize area while minimizing perimeter. We can pretend that instead of a triangle, we are going to maximize the area for a rectangle (since this is easier) and divide our resultant area by two. So, let's maximize the area for the rectangle with sides $y,ky$ where our perimeter must be $2y+2ky \le c$ where $c$ is a constant (we don't need to know what $c$ is). If we maximize that area, then we have also maximized the area of the triangle.

Area for the rectangle is just $A=ky^2$. But, now, we know $2y+2ky=c$, so solving for $y$, we get $y=\frac{c}{2+2k}$.

Now, our area formula looks like this:
$A=k(\frac{c}{2+2k})^2 = \frac{kc^2}{4(k+1)^2}$
Maximizing the area, we get:
$0=\frac{4c^2(k+1)^2-8k(k+1)c^2}{16(k+1)^4}$
$(k+1)^2 = 2k(k+1)$
$k^2=1$
$k = \pm 1$

3. ## Re: MAX volume given array of constraints

(If you really want to, you can show the same thing that for any fixed $x$, the triangle has maximum area at $k=1$, but this is tedious and unnecessary.)

4. ## Re: MAX volume given array of constraints

Okay, I get
$y_{max}=\frac{2}{3+\frac{3}{2}\sqrt{2}}=.3905$ because y>0
Hence,
$V_{max} = \frac{1}{2}(-2(\frac{2}{3+\frac{3}{2}\sqrt{2}})-(\frac{2}{3+\frac{3}{2}\sqrt{2}})\sqrt{2}+2)(\frac {2}{3+\frac{3}{2}\sqrt{2}})^2=0.0508m^2$

5. ## Re: MAX volume given array of constraints

That's what I get, as well.

6. ## Re: MAX volume given array of constraints

Thanks a lot for the help. Can I ask you what promoted you to take this approach? I'd like to think I am okay at math, however, I had been working on this problem for nearly 4 hours with no luck, and I certainly would not have thought about doing that.

7. ## Re: MAX volume given array of constraints

In general, three variables is too many to effectively maximize. So, I needed to simplify the problem. My intuition told me that an isosceles triangle would yield the maximal area, and that the maximal volume would simply be found by taking the maximal area and multiplying it by the length. So, I began playing with the variables until I could express some in terms of others (in order to satisfy the boundary conditions). Once that was accomplished, the trick to fix $x$ and evaluate the area of the cross sectional triangle is not entirely intuitive. There is the question of whether or not the volume is truly maximized over all possible values of $x$. Due to the linearity of the volume formula, I knew it would work, but the specifics as to why require a bit of measure theory, which is not something that you will need for your course. In general, for the problems that you are given, you can usually assume that you can fix one variable (in terms of the others) and maximize the equation that way.

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