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How do you solve this calculus problem?

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?
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1 Answer

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
Using the formula for the volume of a rectangular prism we have
V = x(9-2x)(12-2x)
   = x3-21x2+108x
Take the derivative of the volume with respect to x
V'(x) = 3x2-42x+108
Let V'(x) = 0
So, 3x2-42x+108 = 0
      x2-14x+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