4. Consider all possible rectangular solids defined by 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ 1 and the vector field given by F = (−x 2 − 4xy)i − 6yzj + 12zk. (a) Calculate the total outwards flux through the 6 sides of the solid. (b) Find value(s) for a and b that yield the maximum flux. (c) Calculate the maximum total flux.

Step-by-step explanation:

By using the Divergence Theorem, the flux equals

∫∫∫ div F dV

= ∫∫∫ ((-2x - 4y) + (-6z) + 12) dV

= ∫(z = 0 to c) ∫(y = 0 to b) ∫(x = 0 to a) (12 - 2x - 4y - 6z) dz dy dx

= ∫(z = 0 to c) ∫(y = 0 to b) (12a - a^2 - 4ay - 6az) dy dx

= ∫(z = 0 to c) (12ab - a^2 b - 2ab^2 - 6abz) dx

= 12abc - a^2 bc - 2ab^2 c - 3abc^2.

So, we need to find the maximum value of

f(a,b,c) = 12abc - a^2 bc - 2ab^2 c - 3abc^2, with a,b,c > 0.

First, we find the critical points of f.

f_a = 12bc - 2abc - 2b^2 c - 3bc^2 = bc(12 - 2a - 2b - 3c)

f_b = 12ac - a^2 c - 4abc - 3ac^2 = ac(12 - a - 4b - 3c)

f_c = 12ab - a^2 b - 2ab^2 - 6abc = ab (12 - a - 2b - 6c).

Setting these equal to 0 (and remembering that a, b, c > 0):

12 - 2a - 2b - 3c = 0

12 - a - 4b - 3c = 0

12 - a - 2b - 6c = 0

Solving this system yields a = 3, b = 3/2, c = 1.

By the Second Derivative Test or otherwise, this can easily be checked to yield the maximum flux.

This maximal flux equals f(3, 3/2, 1) = 27/2

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64