I have the following question:
I have a rectangular box with a volume of $6000$ $cm^3$ made from $2$ kinds of materials. The top and bottom are made of a material costing $0.10$/$cm^3$ and the sides from material costing $0.05$/$cm^3$. Find the dimensions of the box so that the cost of materials is minimized.
(sorry for the lack proper formulas but I have tried a few times to figure out how to create them with no success.)
This is what I have so far:
Let $x, y, z$ = length, width and height
$C= (2xy)(0.10) + (2xz + 2yz)(0.05) = 0.20xy + 0.10xz +0.10yz $
Volume = $xyz= 16000 z = \frac{16000}{xy}$
When I substitute in $z$ into the question I have
$C=0.20xy + 0.10x* 16000\frac{x}{y} + 0.10y 16000/xy $
giving me: $C= 0.20xy + \frac{1600}{y} + \frac{1600}{x}$
When I take the partial derivative I get $Cx = 0.20y - \frac{1600}{x^2}$ $Cy = 0.20x - \frac{1600}{y^2}$
And this is where I am stuck. If I solve for x or y I end up with a $\pm\sqrt{8000/y}$.
Any help with this would be appreciated.
Asuming the formula to be correct, you properly found that $$C'_x=\frac{y}{5}-\frac{1600}{x^2}=0\qquad , \qquad C'_y=\frac{x}{5}-\frac{1600}{y^2}=0$$ Because of the symmetry $x=y$. So $$\frac{x}{5}-\frac{1600}{x^2}=0 \implies x^3=8000\implies x=y=20$$ If you do not see the symmetry, from $C'_x=0$, extract $y$ $$C'_x=0 \implies y=\frac{8000}{x^2}\implies C'_y=\frac{x}{5}-\frac{x^4}{40000}=\frac{x(8000-x^3)}{40000}=0$$
Edit
Since you solved the problem, let us make it more general : the given volume is $V$, for top and bottom the cost is $a$, for the sides the cost is $b$. Doing the same as you did $$C=2 a x y+2 b (x z+y z)=2 a x y+2 b \left(\frac{V}{x}+\frac{V}{y}\right)$$ Differentiating $$C'_x=2 a y-\frac{2 b V}{x^2}=0\qquad , \qquad C'_y=2 a x-\frac{2 b V}{y^2}=0$$ where we see again the symmetry. Doing the same as above, the solution is then given by $$x=y=\sqrt[3]{\frac{b V}{a}}\qquad z=\sqrt[3]{\frac{a^2 V}{b^2}}$$ So, as expected, the base is a square and the dimensions depend on the ratio of the costs of used materials.