Optimization problem-find the rectangular box dimensions

Question

A rectangular box with no top is to be made having volume 12 cubic feet.Cost per sq ft of the material to be used is Tk 4 for the bottom,Tk 3 for two opposite sides and Tk 2 for remaining opposite sides. Find dimensions of the box for minimum cost.

Answer 1

Starting from @brenderson's answer, you need to minimize $$f= 4x_1x_2 + 6x_1x_3 + 4x_2x_3 \quad \text{subject to}\quad x_1x_2x_3 = 12$$ When you face equality constraints, it is often good to eliminate some wriables from the constaints.

In this case, $x_3=\frac {12}{x_1\,x_2}$ makes $$f=4 x_1 x_2+\frac{48}{x_1}+\frac{72}{x_2}$$ Compute the partial derivatives and set them equal to $0$ $$\frac{\partial f}{\partial x_1}=4 x_2-\frac{48}{x_1^2}=0\implies x_2=\frac{12}{x_1^2}$$ $$\frac{\partial f}{\partial x_2}=4 x_1-\frac{72}{x_2^2}=\frac{1}{2} x_1 \left(8-x_1^3\right)=0\implies x_1=2$$ So $x_1=2$ , $x_2=3$ and $x_3=2$

Answer 2

Let the length of the box edges be $x_1,x_2,x_3\in\mathbb{R}$, where $x_3$ is the height. Define $x=(x_1,x_2,x_3)$. Then the cost of the box is \begin{equation} f(x) = 4x_1x_2 + 6x_1x_3 + 4x_2x_3. \end{equation} To minimize the cost of the box, subject to your volumetric constraint, it suffices to solve the following problem: \begin{equation} \begin{aligned} & \text{minimize} && f(x) = 4x_1x_2 + 6x_1x_3 + 4x_2x_3 \\ & \text{subject to} && g(x) = x_1x_2x_3 - 12 = 0. \end{aligned} \end{equation} Define the Lagrange multiplier $\lambda\in\mathbb{R}$ and the Lagrangian \begin{equation} L(x,\lambda) = f(x) + \lambda g(x) = 4x_1x_2 + 6x_1x_3 + 4x_2x_3 + \lambda(x_1x_2x_3-12). \end{equation} Then the Lagrangian stationarity KKT condition requires that \begin{equation} \nabla_x L(x^*,\lambda^*) = \begin{bmatrix} 4x_2^* + 6x_3^* + \lambda^* x_2^*x_3^* \\ 4x_1^* + 4x_3^* + \lambda^* x_1^*x_3^* \\ 6x_1^* + 4x_2^* + \lambda^* x_1^*x_2^* \end{bmatrix} = 0. \end{equation} Therefore, by multiplying these three equations by $x_1^*$, $x_2^*$, and $x_3^*$, respectively, we may substitute the primal feasibility KKT condition (namely, $x_1^*x_2^*x_3^*=12$) to obtain \begin{equation} \begin{bmatrix} 4x_1^*x_2^* + 6x_1^*x_3^*+12\lambda^* \\ 4x_1^*x_2^* + 4x_2^*x_3^*+12\lambda^* \\ 6x_1^*x_3^* + 4x_2^*x_3^* + 12\lambda^* \end{bmatrix} = 0. \end{equation} Subtracting the second equation from the first, we find that $6x_1^*x_3^*-4x_2^*x_3^*=0$, which has the nontrivial solution $x_1^*=\frac{2}{3}x_2^*$. Subtracting the third equation from the second, we find that $4x_1^*x_2^*-6x_1^*x_3^*=0$, which has the nontrivial solution $x_3^*=\frac{2}{3}x_2^*$. Therefore, substituting these results back into the primal feasibility condition, we find that \begin{equation} x_1^*x_2^*x_3^*=\frac{4}{9}x_2^{*3}=12, \end{equation} which implies that \begin{equation} x^*=(x_1^*,x_2^*,x_3^*)=(2,3,2). \end{equation} Note that plugging this primal solution back into the Lagrangian stationarity condition yields $\lambda^*=-4$ from all three equations.