Pick a point $A = (x,y,z)$ on the plane so that $x + 3y + 6z = 18$.
The volume $V$ of the box is $V = xyz$. So we find $max(V)$ with conditions $x + 3y + 6z = 18$ and $x, y, z \in \mathbb{R}^{\text{nonneg}}$.
Use $g(x,y,z) = x + 3y + 6z = 18$. So
$$V'(x) = yz = rg'(x) = r$$
$$V'(y) = xz = rg'(y) = 3r$$
$$V'(z) = xy = rg'(z) = 6r$$
This means $xz = 3yz , xy = 6yz \Rightarrow z(x - 3y) = 0, y(x - 6z) = 0$.
If $z = 0$ or $y = 0$, then $V = 0$ and is not a max. So $$x - 3y = 0 = x - 6z \Rightarrow y = \frac{x}{3}, z = \frac{x}{6}$$
$$\Rightarrow x + 3\left(\frac{x}{3}\right) + 6\left(\frac{x}{6}\right) = 18$$
$$\Rightarrow 3x = 18 \Rightarrow x = 6, y = 2, z = 1$$ So $max(V) = 6*2*1 = 12$.
Let $x$ be the size of the base and $y$ be the height of the box. The volume is $v=x^2y$. The cost is $ C=75x^2+200xy+20x^2=95x^2+200xy.$ Therefore we have to maximize $x^2y,$ subject to the constraint $95x^2+200xy=400.$ We solve for y as a function of $x$ to get $$ y=(400-95x^2)/200x$$ Upon simplifying and substituting in the formula for $v$ we find $$v=x^2(400-95x^2)/200x =2x-(19/40)x^3.$$ $v'=0$ implies $x=\sqrt {80/57}.$ substituting in the expression for $y$ we get $y=19\sqrt {57/5}.$ therefore the maximum volume is $$v_{max} =1444\sqrt {57/5}$$
Best Answer
Constraint equation: $x+2y+2z = 100$, in order to obtain maximum volume. Also $V=x \cdot y \cdot z$. Solving for $x$ we get $x=100-2y-2z$, substitute into our volume equation to obtain $V=(100-2y-2z) \cdot y \cdot z$.
$V_y=100z-4yz-2z^2$
$V_z=100y-2y^2-4yz$.
Now we want to solve $V_y=0$ and $V_z=0$. Hence,
$100z-4yz-2z^2=0$ ... (1)
$100y-2y^2-4yz=0$ ... (2)
From equation (2) we get $y=50-2z$, substituting this into equation (1) we get $z=\frac{50}{3}$ and from the comments, due to symmetry, we get $y=\frac{50}{3}$. We can now solve for $x$ to get $x=\frac{100}{3}$.
Therefore, Length=$\frac{100}{3}$in, Width=$\frac{50}{3}$in, Height=$\frac{50}{3}$in.