Answer
\[
\left.f\right|_{(2,2,4)}=240 \quad \text { cents }
\]
Work Step by Step
We know:
\[
\begin{array}{c}
f(x, y)=(2 x y)10+(2 y z+2 z x)5 \\
20 x y+10 y z+10 z x=f(x, y)
\end{array}
\]
The constraint is $-16+x y z=g(x, y)$. We want to maximize the function $f$ by using Lagrange multipliers:
\[
\lambda \nabla g=\nabla f
\]
We will need the gradient
\[
\begin{aligned}
\nabla f &=20 y \hat{i}+10 z \hat{j}+10 x \hat{k} \\
\nabla g &=y \hat{z}+z x \hat{j}+x y \hat{k} \\
\because \quad \nabla f &=\lambda \nabla g \\
(20 y \hat{i}+10 z \hat{j}+10 x \hat{k}) &=\lambda(y z \hat{i}+z x \hat{j}+x y \hat{k})
\end{aligned}
\]
Which is equivalent to the pair of equations
\[
\begin{aligned}
&\lambda(y z) =20 y \\
&\lambda(z x)=10z \\
&\lambda(x y)=10x \\
\Rightarrow \frac{20 y}{y z} &=\frac{10 z}{z x}=\frac{10 x}{x y}
\end{aligned}
\]
Thus, from equation(1), $z=2 y=2 x,$ so:
$-16+x y z$
\[
16=2 x^{3}, \quad \Rightarrow 2=x
\]
\[
\begin{array}{l}
\left.f\right|_{(2,2,4)}=20(2 \times 2)+10(2 \times 4)+10(2 \times 4) \\
\left.f\right|_{(2,2,4)}=240
\end{array}
\]
