Minimizing $c(x-2)(y-2)(z-2)$ subject to $xyz =V$ Given constants $c > 0$ and $V > 0$, how can I solve the following optimization problem?
$$\begin{array}{ll} 
\text{minimize} & c \, (x-2) (y-2) (z-2)\\ 
\text{subject to} & xyz = V\\ 
& x,y,z > 2
\end{array}$$
My attempt
Since $V=xyz$, we get $z = \frac{V}{xy}$
Let $S = c(x-2)(y-2)(z-2) = c(x-2)(y-2)(\frac{V}{xy}-2)$
Simplifying we get,
$$S = c(4x+4y-2xy-2\frac{V}{x}-2\frac{V}{y}+4\frac{V}{xy}+V-8)$$
Partially differentiate $S$ w.r.t. $x$ and $y$ and equate to $0$.
$\frac{\partial S}{\partial x} = c (4 -2y + \frac{2V}{x^{2}} - 4\frac{V}{x^{2}y}) = 0$ -------- (A) 
$\frac{\partial S}{\partial y} = c (4 -2x + \frac{2V}{y^{2}} - 4\frac{V}{y^{2}x}) = 0$ -------- (B)
If we subtract (B) from (A), then one solution is $x=y$. 
Is it correct till this point ?
 A: There is NO minimum.  If you allowed x, y, and z equal to 2, then the minimum would be for x= y= 2 and z= V/4.  However, requiring that x, y, and z all be larger than 2, you can get as close to that value as you wish by taking x and y as close to 2 as you wish.
A: Let$f(x,y,z):=c (x - 2) (y - 2) (z - 2)$, and $g(x,y,z)=x y z - V$. Then solve for $x,y,z,\lambda$ in:
$$\frac{\partial f(x,y,z)}{\partial x}=\lambda  \frac{\partial g(x,y,z)}{\partial x}\\
\frac{\partial f(x,y,z)}{\partial y}=\lambda  \frac{\partial g(x,y,z)}{\partial y}\\\frac{\partial f(x,y,z)}{\partial z}=\lambda  \frac{\partial g(x,y,z)}{\partial z}\\g(x,y,z)=0$$
The complete solution set should be:
$$\left\{x\to 2,y\to 2,z\to \frac{V}{4},\lambda \to 0\right\},\left\{x\to 2,y\to \frac{V}{4},z\to 2,\lambda \to 0\right\},\color{red}{\left\{x\to \sqrt[3]{V},y\to \sqrt[3]{V},z\to \sqrt[3]{V},\lambda \to \frac{c \left(\sqrt[3]{V}-2\right)^2}{V^{2/3}}\right\}}\\
\left\{x\to -\sqrt[3]{-1} \sqrt[3]{V},y\to -\sqrt[3]{-1} \sqrt[3]{V},z\to -\sqrt[3]{-1} \sqrt[3]{V},\lambda \to c \left(1-\frac{4 \left((-1)^{2/3} \sqrt[3]{V}+\sqrt[3]{-1}\right)}{V^{2/3}}\right)\right\},\left\{x\to (-1)^{2/3} \sqrt[3]{V},y\to (-1)^{2/3} \sqrt[3]{V},z\to (-1)^{2/3} \sqrt[3]{V},\lambda \to \frac{c \left(\sqrt[3]{V}+2 \sqrt[3]{-1}\right)^2}{V^{2/3}}\right\},\left\{x\to \frac{V}{4},y\to 2,z\to 2,\lambda \to 0\right\}$$
Since the $x,y,z>2$ and $V,c>0$, then only the answer in red is meaningful.

Following your approach for $z=\frac V{x y}$,for $h(x)=c (x-2) (y-2) \left(\frac{V}{x y}-2\right)$:
$$\frac{\partial h(x,y)}{\partial x}=0\\
\frac{\partial h(x,y)}{\partial y}=0
$$
Your complete solution set should be:
$$\left\{\left\{x\to 2,y\to \frac{V}{4}\right\},\color{red}{\left\{x\to \sqrt[3]{V},y\to \sqrt[3]{V}\right\}},\left\{x\to -\sqrt[3]{-1} \sqrt[3]{V},y\to -\sqrt[3]{-1} \sqrt[3]{V}\right\},\left\{x\to (-1)^{2/3} \sqrt[3]{V},y\to (-1)^{2/3} \sqrt[3]{V}\right\},\{x\to 2,y\to 2\},\left\{x\to \frac{V}{4},y\to 2\right\}\right\}$$
