Pls I’ll literally give brainliest to whoever

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.

7 0

3 years ago

‼️WILL LIST BRAINLIEST ‼️Explain the Error: Jayme tells her friend that ordered pairs that have an x-coordinate of O lie on the

Sergio [31]

Because if they were ordered pairs they would have to to not all of them could have an x value of 0 and still lie on the x axis

4 0

3 years ago

23.) Clarissa earns $300 per

Talja [164]

Answer:

e = 300 + .0275c

Step-by-step explanation:

So the earnings = e

Then she earns $300 and a bonus for cosmetics : 2.75%c which is equal to 0.0275 c or .0275c the c for cosmetics.

Thus the answer becomes :

e = $300 + .0275c

HOPE THIS HELPED

3 0

3 years ago

The system of equations x + 2y = 6 and x – 4y = 0 is graphed below. What is the solution to the system of equations?

blagie [28]

Step-by-step explanation:

$x + 2y = 6....(1) \\ \\ x - 4y = 0\\ \implies \: x = 4y....(2)\\ \\ from \: equations \: (1) \: and \: (2) \\ 4y + 2y = 6 \\ \therefore \: 6y = 6 \\ \\ \therefore \: y = \frac{6}{6} \\ \\ \huge \red{ \boxed{ \therefore \: y = 1}} \\ \\ substituting \: y = 1 \: in \: equation \: (2) \\ \\ x = 4y \\ \\ \therefore \: x = 4 \times 1 \\ \\ \huge \orange{ \boxed{\therefore \: x = 4 }} \\ \\ \therefore \: (x, \: \: y) = (4, \: \: 1) \\ is \: the \: required \: solution. \\$