Question

a cargo container (in the shape of a rectangular solid) must have a volume of 400 cubic feet. the bottom will cost $5 per square foot to construct and the sides and the top will cost $3 per square foot to construct. use lagrange multipliers to find the dimensions of the container of this size that has minimum cost.

Explanation:

Step1: Define variables and functions

Let the length be \(x\), width be \(y\), and height be \(z\). The volume \(V = xyz=400\). The cost function \(C = 5xy+3(2xz + 2yz)=5xy + 6xz+6yz\). We use the Lagrange function \(L(x,y,z,\lambda)=5xy + 6xz+6yz-\lambda(xyz - 400)\).

Step2: Calculate partial derivatives

\(\frac{\partial L}{\partial x}=5y + 6z-\lambda yz = 0\), so \(\lambda=\frac{5y + 6z}{yz}=\frac{5}{z}+\frac{6}{y}\).
\(\frac{\partial L}{\partial y}=5x + 6z-\lambda xz = 0\), so \(\lambda=\frac{5x + 6z}{xz}=\frac{5}{z}+\frac{6}{x}\).
\(\frac{\partial L}{\partial z}=6x + 6y-\lambda xy = 0\), so \(\lambda=\frac{6x + 6y}{xy}=\frac{6}{y}+\frac{6}{x}\).
\(\frac{\partial L}{\partial\lambda}=-(xyz - 400)=0\), so \(xyz = 400\).

Step3: Solve the system of equations

From \(\frac{5}{z}+\frac{6}{y}=\frac{5}{z}+\frac{6}{x}\), we get \(x = y\).
Substitute \(x = y\) into \(\frac{5}{z}+\frac{6}{x}=\frac{6}{x}+\frac{6}{x}\), we have \(5x=6z\), so \(z=\frac{5}{6}x\).
Substitute \(y = x\) and \(z=\frac{5}{6}x\) into \(xyz = 400\), we get \(x\cdot x\cdot\frac{5}{6}x=400\), \(x^{3}=480\), \(x = y=\sqrt[3]{480}\approx7.83\). Then \(z=\frac{5}{6}\sqrt[3]{480}\approx6.52\).

Answer:

The length and width are approximately \(7.83\) feet and the height is approximately \(6.52\) feet.