Multi-Variable Calculus Question

@ = my computer symbol for partial

Okay.

"The material for constructing the base of an open box costs 1.5 times as much per unit area as the material to construct the sides. For a fixed amount of money C, find the dimensions of the box of largest volume that can be built."

V = xyz
C = 1.5xy + 2xz + 2yz
^ derived to
z = (C-1.5xy)/(2x+2y)

I've already found @V/@x and @V/@y, and x = y, as expected. However, I cannot for the life of me find out how to get x or y in terms of C, which I assume is what the problem is asking for.

Help?

I find x = y = 4*sqrt(C)/(6*sqrt(2)), and z = 3*sqrt(C)/(6*sqrt(2)), which probably doesn't help. xD

Anyway, were you taught about Lagrangian multiplicators/functions?

Quote:

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Originally Posted by Endless

I find x = y = 4*sqrt(C)/(6*sqrt(2)), and z = 3*sqrt(C)/(6*sqrt(2)), which probably doesn't help. xD

Anyway, were you taught about Lagrangian multiplicators/functions?

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He taught those today, after this was due. Something like Gradient of f(x,y) = Lambda gradient g(x,y), right? I didn't really pay attention, because I worked crossword puzzles the whole period.

I eventually did something here or there, and came up with

X = Y = (Root2C)/3
Z = C/(2root2C)

Several other people came up with the same answer.

We have the same answers, I just didn't simplify 4/(6*sqrt(2)) into sqrt(2)/3 and 3/(6*sqrt(2)) into 1/(2*sqrt(2)), and you didn't siplify C/sqrt(C) into sqrt(C).