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The Box
This is a common problem often encountered in basic algebra and differential calculus in the study of general cubic and quadratic
polynomial equations. It involves finding the maximum possible volume of a box cut out
from a piece of rectangular cardboard with dimensions (L x W) where (L) and (W) are the
length and width. The numerical values can represent any convenient units of measure, inches, cm, feet, meters, etc..
The standard quadratic equation and its solution looks like:
We often see lessons showing the solution to standard quadratic equations, but where the coefficient values A, B and C come from and what they mean is not always clear. A quadratic equation does not only apply to parabolic curves. This practical example shows the creation of a quadratic equation derived as the first derivative of a cubic volume equation, demonstrating one means by which the quadratic coefficients A, B and C originate and also the general solution to the resulting equation.
Quadratic equations are fundamental to multiple fields, such as mathematics, statistics, electronics and engineering, the physical sciences and more. Even some problems in relativity can be reduced to quadratic solutions.
CARDBOARD SHEET WITH DIMENSIONS (L × W) TO BE CUT
AND FOLDED INTO A BOX OF MAXIMUM POSSIBLE VOLUME
Below we will derive the specific equation needed to solve ALL
maximum volume box problems of this type in general in terms
of L, W and x.
Given the piece of cardboard with dimensions (L × W), we cut out 4 identical squares with edges of length (x) from each corner of the cardboard as shown below. The value of (x) will be the depth of the resulting box. However, at this point, we do not yet know what that optimum value of (x) should be to solve the maximum volume problem. By varying (x), an indefinite number of boxes could be created from the cardboard, but only one value of (x) solves the problem at hand.
The problem is to find the optimum value (x) of the lengths of the sides of the squares that need to be cut out of the cardboard to yield the maximum possible volume (V), using the source cardboard with the dimensions given above.
The cardboard would look something like this after cutting out the corners, but we still have to find that special optimal value of (x) for maximum volume. We can now see what will be the sides of the box.
Now, the sides could then be folded upward to form a box with the dimensions and volume (V), similar to that shown here.
Now, we can proceed to find the optimum value (x) of the lengths of the sides of the squares that need to be cut out of the cardboard to yield the maximum possible box volume (V).
To do this, first expand and express (V) in the form of a cubic polynomial equation:
Next, we find the first derivative of the volume (V) with respect to (x) which results in a quadratic equation which we equate to zero to find the optimum value of (x) that yields the maximum possible volume (V).
This gives the extremum (in this case, maximum) value. The solution (x) to the quadratic equation is the value we seek. The quadratic equation has two solution roots, but only one of the (x) values is the correct solution to the physical problem at hand. It is not difficult to determine which value applies.
In the form of a standard quadratic equation, the elements are:
Substituting the expressions for A, B and C and reducing the quadratic equation to its simplest form yields the general solution for (x) purely in terms of (L) and (W). The quadratic equation will have two solutions for (x), of which only one will apply to the physical problem.
Now, solving for the two quadratic roots gives:
 = 14.724747683481 units
and
 = 45.275252316519 units
Of the above two quadratic (x) roots, only (x1) applies. The alternate root (x2) can be ignored as physically inapplicable to the problem. Calling the applicable (x1) value simply (x), the final general solution we seek is:
 = 14.724747683481 units
Thus, this final equation for (x) solves the general case of the maximum volume box problem given the (L) and (W) dimensions of the cardboard or whatever material is used, ignoring the thickness of the material.
Given the value of (x), the volume of the box can be found from:
 = 52513.80432417 cubic units
Summary for this case:
L = Length of original cardboard sheet = 100 units
W = Width of original cardboard sheet = 80 units
A = Area of bottom of created box = 8 square units
x1 = Optimum (x) Value = Box Depth = 14.724747683481 units
V1 = Maximum possible volume of the folded box = 52513.80432417 cubic units
The alternate (inapplicable) values are:
x2 = 45.275252316519 units
V2 = -4513.8043241697 cubic units
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