SOLUTION: A box is to be constructed from a sheet of cardboard that is 20 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides. What is the maxim

Question 985067: A box is to be constructed from a sheet of cardboard that is 20 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides.
What is the maximum volume this box could have? (Round your answer to two decimal places. Do not include units, for example, 10.22 cm would be 10.22.)
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
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L=length=50-2x; W=width=20-2x; H=height=x; V=volume
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The function will have a maximum at the point where the first derivative equals zero.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1900 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 18.9314982392345, 4.40183509409888. Here's your graph:

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Determine the domain of the function. x must be greater than 0 or H would be zero and we would have a flat sheet, so x>0. 2x must be less than 20 or we would have no width so 2x<20 or x<10
So 0 < x < 10.
That leaves us with x=4.40 as the solution.
The size of the box:
L=50-2x=50-8.80=41.20
W=20-2x=20-8.80=11.20
H=x=4.40
Maximum Volume of the box=(41.20)(11.20)(4.40)=2030.34 cm^3

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