Step Two: Identify the constraint.
The constraint is the rule or equation that relates the variables used to generate the
objective function. In this case, the way to relate the variables x and y is to use the fact that the total
price of the box materials must equal $20. Since the cost of the material is the area of
the material multiplied by the cost per square foot, the constraint can be expressed as
follows:
(4xy)(2) + (x^{2})(4) = 20
Step Three: Use the constraint to express the objective as a function of one
variable.
The methods that we have learned to analyze functions only apply to functions of one
variable. The constraint can be used to reduce the objective to a function of one variable
so that our techniques of finding maxima and minima will apply. This involves using the constraint to solve for one variable
in terms of another. In this case, we solve for y, although solving for x will work also:
y = = 
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Now, this can be substituted back into the original objective to yield:
Step Four: Now, V is expressed as a function of one variable, x, and
procedures explained previously to for optimizing functions of one variable can be used.
The domain of V(x) is (0, + ∞). This is because x could never be a negative
quantity, and could not be zero.
V'(x)   =  x^{2} 

V'(x)   = 0 whenx = ± 

but only x = + is in the domain of V.
Now, to check if this critical point is a local maximum, minimum, or neither, the second
derivative test can be used:
  V''(x) =  3x 

  V'' =  3 < 0 

Because the second derivative is negative, this critical point is a local maximum.
We can also be sure that this is the absolute maximum on the open interval (0, + ∞).
This is because there are there are no more critical points on this interval, so the graph
must only be increasing to the left of the critical point, and decreasing to the right. To
answer the original problem, the largest possible volume is: