Answer to Question #289071 in Calculus for Kang

Question

A factory produces a closed rectangular parallelepiped vats having the capacity of 10 cubic meters. Find the dimensions that will make the cost of the lining a minimum?

Answer 1

The volume of a rectangular parallelepiped is $xyz$ which is 10 cubic meters

$$xyz=10$$

The surface area of the figure is $2xy+2xz+2yz$ which is to be minimized.

Using Langrage method.

$$F(x,y,z,\lambda)=(2xy+2xz+2yz)+\lambda(xyz-10)\\\\ \frac{\partial F}{\partial x}=2y+2z+\lambda yz\\\\ \frac{\partial F}{\partial y}=2x+2z+\lambda xz\\\\ \frac{\partial F}{\partial z}=2x+2y+\lambda xy\\ \frac{\partial F}{\partial \lambda}=xyz-10\\$$

Equating the partial derivatives to zero.

$$2y+2z+\lambda yz=0~~~~~~~~~~~~~~~~~~~(1)\\ 2x+2z+\lambda xz=0~~~~~~~~~~~~~~~~~~~(2)\\ 2x+2y+\lambda xy=0~~~~~~~~~~~~~~~~~~~(3)\\ xyz-10=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)$$

Solving (1) and (2) together, we have that;

$$x=y$$

Also, solving (2) and (3) together, we have that;

$$y=z$$

Thus,

$$x=y=z$$

Substituting this into (4), we have that;

$$x^3-10=0\\ x^3=10\\ x=\sqrt[3]{10}$$

Hence the dimension to minimize the lining is

$$\sqrt[3]{10}\times \sqrt[3]{10}\times \sqrt[3]{10}$$