SOLUTION: An open-top box is to be constructed from a 6 foot by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the lengt

Question 63266: An open-top box is to be constructed from a 6 foot by 8 foot rectangular cardboard by cutting out equal squares at each corner and the folding up the flaps. Let x denote the length of each side of the square to be cut out. Find the function V that represents the volume of the box in terms of x. Graph this fuction and show the graph over the valid range of the variable x. Using the graph, what is the value of x that will produce the maximum volum?
Answer by jai_kos(139) (Show Source): You can put this solution on YOUR website!

Imagine the 6 by 8 cardboard with squares cut out of it.
|-------6---------|
...._______
__|.............|__ ....__
|.x............... x | .....|
|......................| .....|
|......................| .....| 8
|__..............__| .....|
..x..|______| x ..._|_
When you fold it up, you have a cube of have the height is x, the width is 6-2x, and the length is 8 - 2x.
The volume is, (6-2x)(8-2x)x so ...
V = (6-2x)(8-2x)x = 4x^3 - 28x^2 +48x
Let us first find the domain of x. x has to be bigger than 0, because it is a length, and less than 3, because otherwise the side 6-2x would be negative. So the domain of x is 0 We find that the maximum volume occurs at x = about 1.1315, using a graph. This creates a volume of about 24.2584
We can calculate this value exactly using algebra as well. We realize that the max value must be a double root of the equation 4x^3 - 28x^2 +48x + D for some value of D.
We do the algebra, and find out that this means that
x = 7/3 - √(13)/3, and at this value, the volume is (104√13)/27 + 280/27. These values do approximate to the approximations above.

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