An open rectangular box is to be made from a 9X12 piece of tin by cutting squares of side x from the corners and folding up the sides. What should X be to maximize the volume of the box?

Here, If we visualize the problem, when we fold the sides up, x becomes the height. And sides 12 and 9 lose 2 of x inches from each side. And volume is l*w*h. In this case, the cuts have to be 0<x<4.5 because of side 9. So, after we cut out the squares of side x and fold up the sides, the dimensions of the box will be:

width: 9-2x

length: 12-2x

height (depth): x

length: 12-2x

height (depth): x

Using the formula for the volume of a rectangular prism we have

V = x(9-2x)(12-2x)

V = x(9-2x)(12-2x)

= x

^{3}-21x^{2}+108xTake the derivative of the volume with respect to x

V'(x) = 3x

^{2}-42x+108Let V'(x) = 0

So, 3x

^{2}-42x+108 = 0 x

^{2}-14x+36 = 0 [Divide by 3]Now solving for x following quadratic formula we get,

x = 3.39 and x = 10.61

Since, 10.61 is not between 0 and 4.5, the valid value of x= 3.39

**Hence, to maximize the volume of the box, x should be 3.39**