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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.

Mark Fantini
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user191919
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  • Is the box of a general shape or rectangular? If it is the latter then front implies back and lateral faces imply each other by symmetry. After clarifying that I would look at lagrangian multipliers. – Nox Feb 08 '14 at 12:43
  • Yes, its rectangular. What do you mean by the front imply the back. I could still glue two pieces on the front and just one piece at the back. – user191919 Feb 08 '14 at 12:54

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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 @ $\mathbf{x}^*$.

Max Herrmann
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