Optimization problem formulation The problem asks me to formulate the following optimization problem, find the first order conditions, solve them and verify second order conditions. So the problem is the following. I have to maximize the volume of a cardboard box, given that the front face, the top and the bottom need to have two pieces each. The constraint on the box is that I have 70 sq ft. of cardboard.
My attempt. Let $x$ denote the width, $z$ the height and $y$ the length.
$$
\begin{align}
\operatorname{max}\limits_{x,y,z} & \quad xyz \\
 \operatorname{s.t.} & \quad \underbrace{xz+xz}_{\text{front face}} + \underbrace{xy+xy}_{\text{top}} + \underbrace{xy+xy}_{\text{bottom}} \\
 & \quad + \underbrace{2zy}_{\text{lateral faces}} + \underbrace{xz}_{\text{back face}} = \, 70.
\end{align}
$$
My problem: the questions asks to solve (with no mention to software) the first order conditions. With the way I modeled it, I have an intractable system of 4 equations to solve manually. So I think the problem is in the way I modeled it; I was hoping someone could shed a light on that.
Thanks in advance.
 A: You modelled it correctly.
$1^{\text{st}}$-order necessary optimality conditions are $(\nabla \bar{f})(\mathbf{x}^*)= 0$, where $\bar{f}:\mathbb{R}^2 \rightarrow \mathbb{R}$ is the reduced cost function after the decision variable $x$ was substituted by ($A$ is cardboard area)
$$ x = \frac{A - 2yz}{3z + 4y}, $$
due to the equality constraint, giving rise to the unconstrained optimality problem
$$ \min_{(y,z) \in \mathbb{R}^2} \frac{A - 2yz}{3z + 4y}yz. $$
Analytic derivation yields the following set of algebraic equations
$$
\begin{align}
\partial_y \bar{f}=-2yz^2(3z+4y)-4(A-2yz)yz + (A-2yz)(3z+4y)z = 0 \\
\partial_z \bar{f}=-2y^2z(3z+4y)-3(A-2yz)yz + (A-2yz)(3z+4y)y = 0 \\
\end{align}
$$
which can be solved numerically ($\texttt{fsolve}$/Octave) returning ($A = 70$)
$$ (y^*,z^*) = (2.958, 3.9441). $$
The triple $\mathbf{x}^* = (1.972, 2.958, 3.9441)$ evaluates the Hessian $\Delta \bar{f}$ negative definite, satisfying the $2^{\text{nd}}$-order sufficient condition for a maximum @ $\mathbf{x}^*$.
