The question for this is "Kiera plays a board game with a spinner like the one shown below.

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

8 0

The answer is C
This is because the Go Back space is twice as big as the Return To Start space

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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 ≤

dusya [7]

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

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

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

We can see that $g$ is quadratic in $a$ , so let's complete the square.

$g(a,b,c)=-3bc(a^2-10a+25-25)=3bc(25-(a-5)^2)$

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