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: Set up the cost and volume functions

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

Step2: Use Lagrange multipliers

Define the Lagrangian function \(L(x,y,z,\lambda)=5xy + 6xz+6yz-\lambda(xyz - 400)\).
Take the partial derivatives:
\(\frac{\partial L}{\partial x}=5y + 6z-\lambda yz = 0\) (1)
\(\frac{\partial L}{\partial y}=5x + 6z-\lambda xz = 0\) (2)
\(\frac{\partial L}{\partial z}=6x + 6y-\lambda xy = 0\) (3)
\(\frac{\partial L}{\partial\lambda}=-(xyz - 400)=0\) (4)
From (1) and (2):
\(5y + 6z-\lambda yz=5x + 6z-\lambda xz\)
\(5y-\lambda yz=5x-\lambda xz\)
\(5(y - x)=\lambda z(y - x)\)
Case 1: If \(y
eq x\), then \(\lambda z = 5\).
From (1) and (3):
From (1) \(\lambda=\frac{5y + 6z}{yz}\), from (3) \(\lambda=\frac{6x + 6y}{xy}\)
If we assume \(x = y\) (since if \(y
eq x\) leads to a contradiction in the non - trivial case for minimizing cost).
Substitute \(x = y\) into (1): \(5x+6z-\lambda xz = 0\), substitute \(x = y\) into (3): \(12x-\lambda x^{2}=0\), so \(\lambda=\frac{12}{x}\)
Substitute \(\lambda=\frac{12}{x}\) into \(5x + 6z-\lambda xz = 0\): \(5x+6z-12z=0\), \(5x = 6z\), \(z=\frac{5}{6}x\)
Substitute \(y = x\) and \(z=\frac{5}{6}x\) into \(xyz = 400\):
\(x\cdot x\cdot\frac{5}{6}x=400\)
\(\frac{5}{6}x^{3}=400\)
\(x^{3}=480\)
\(x = y=\sqrt[3]{480}\approx7.83\)
\(z=\frac{5}{6}\sqrt[3]{480}\approx6.52\)

Answer:

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