Question

A box is contructed out of two different types of metal. the metal for the top and bottom, which are both square, costs $3 per square foot and the metal for the sides costs $10 per square foot. find the dimensions that minimize cost if the box has a volume of 15 cubic feet.

Answer 1

Let x ft be the length of the base square and y ft be the height of the box.

The volume of the box is

$V=x\cdot x\cdot y\ ft^3.$

Since the box has a volume of 15 cubic feet, then

$x^2y=15,\\ \\y=\dfrac{15}{x^2}.$

You need to construct two squares from the metal that cost $3 per square foot.

The area of each square is $x^2\ ft^2$ and the total cost for these two squares is $2\cdot x^2\cdot 3=6x^2.$

The area of each side face is $x\cdot y=x\cdot \dfrac{15}{x^2}=\dfrac{15}{x}.$ Then the total cost for sides is $4\cdot \dfrac{15}{x}\cdot 10=\dfrac{600}{x}.$

Let S(x) be the function that represents total cost of the box, then

$S(x)=6x^2+\dfrac{600}{x}.$

Find the derivative:

$S'(x)=12x-\dfrac{600}{x^2}.$

When $S'(x)=0,$ then $12x-\dfrac{600}{x^2}=0,\\ \\12x^3=600,\\ \\x^3=50,\\ \\x=\sqrt[3]{50}\ ft.$

The dimensions of the box are:

length and width - $\sqrt[3]{50}\ ft$

height - $\dfrac{15}{\sqrt[3]{2500}}\ ft.$