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Find the dimensions of the rectangular box with largest volume if the total surface area is given as $64$cm$^2$

My approach:

$$ v=f(x,y,z)= xyz\\ 2xy+2yz+2xz = 64\\ xy+yz+xz = 32 $$

Then I solved for $xy$: $$ xy = 32 - yz - xz $$ Substituted back in $v$: $$ f(x,y,z) = z(32 - yz - xz) = 32z - yz^2-xz^2 $$ Then, I tried to find critical points for that function, to find where I could get the maximum volume: $$ f_x = -z^2\\ f_y = -z^2\\ f_z = 32-2zy-2zx\\ $$

And now I'm lost, because if z=0 from $f_x$ and $f_y$, I can't get $f_z=0$...

What have I done wrong? Do I need to solve for just one variable and then substitute? What's going on here? I really don't get why it's "incorrect" to solve for two variables since my "main" function (in this case $v$) has these two variables in it...

EDIT: $$ xy = 32z - yz - xz $$

is not solved in terms of $xy$, since my right side of equation still has x and y... that's what I have done wrong?!

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Try $xy+xz=32-yz \Rightarrow x=\frac {32-yz}{y+z}$ to get the volume as a function in terms of $y$ and $z$

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  • $\begingroup$ I've done that way and it worked... I just wanna get why my substition did not work. But I think I've figured it out... Check my EDIT to see if what i've said is indeed correct! $\endgroup$ – Bruno Reis Nov 1 '16 at 14:18
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By the symmetry of your problem the solution must be $x=y=z$

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  • $\begingroup$ This is a nice way to approach the problem - but your answer is incomplete in two ways. First, you don't explain why the solution must be symmetric. Second, the OP wants to know what s/he did wrong. $\endgroup$ – Ethan Bolker Nov 1 '16 at 15:56

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