Calculus AB: Applications of the Derivative: Optimization

Calculus AB: Applications of the DerivativeOptimization

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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) + (x²)(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 = = - <BR>

Now, this can be substituted back into the original objective to yield:

V = x²

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)		= - x²
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: