Q:

Let divf = 6(5 − x) and 0 ≤ a, b, c ≤ 12. (a) find the flux of f out of the rectangular solid 0 ≤ x ≤ a, 0 ≤ y ≤ b, and 0 ≤ z ≤c. your answer will be in terms of a, b,c. flux = (b) for what values of a, b, c is the flux the largest? what is that largest flux?

Accepted Solution

A:
Continuing from the setup in the question linked above (and using the same symbols/variables), we have

[tex]\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}(\nabla\cdot f)\,\mathrm dV[/tex]
[tex]=\displaystyle6\int_{z=0}^{z=c}\int_{y=0}^{y=b}\int_{x=0}^{x=a}(5-x)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]
[tex]=\displaystyle6bc\int_0^a(5-x)\,\mathrm dx[/tex]
[tex]=6bc\left(5a-\dfrac{a^2}2\right)=3abc(10-a)[/tex]

The next part of the question asks to maximize this result - our target function which we'll call [tex]g(a,b,c)=3abc(10-a)[/tex] - subject to [tex]0\le a,b,c\le12[/tex].

We can see that [tex]g[/tex] is quadratic in [tex]a[/tex], so let's complete the square.

[tex]g(a,b,c)=-3bc(a^2-10a+25-25)=3bc(25-(a-5)^2)[/tex]

Since [tex]b,c[/tex] are non-negative, it stands to reason that the total product will be maximized if [tex]a-5[/tex] vanishes because [tex]25-(a-5)^2[/tex] is a parabola with its vertex (a maximum) at (5, 25). Setting [tex]a=5[/tex], it's clear that the maximum of [tex]g[/tex] will then be attained when [tex]b,c[/tex] are largest, so the largest flux will be attained at [tex](a,b,c)=(5,12,12)[/tex], which gives a flux of 10,800.