Question

Find the number of units \(x\) that produces a maximum revenue \(R\). $$ R=48 x^{2}-0.02 x^{3} $$

Short answer

The number of units that produces the maximum revenue is 1600.

Step-by-step solution

Calculate the Derivative of \(R(x)\)

The first step is to calculate the derivative of the given function. The derivative of a polynomial function can be calculated using basic differentiation rule i.e. if \(f(x) = ax^n\), then \(f'(x) = n*ax^{n-1}\). By applying this rule, the derivative \(R'(x)\) of \(R(x) = 48x^2 - 0.02x^3\) is \(R'(x) = 96x - 0.06x^2\).

Find the Critical Points

The next step is to find the critical points of the function. These are values of \(x\) where the derivative is zero. So we set the derivative equal to zero and solve for \(x\). That is, solve \(96x - 0.06x^2 = 0\). By factoring out \(x\), we obtain \(x(96 - 0.06x) = 0\), from which we find that \(x = 0\) or \(x = 1600\).

Test the Critical Points

At this point, we test our critical points to see if they result in a local maximum, local minimum, or neither. We can use the second derivative test for this. Compute the second derivative of the function, which is \(R''(x) = 96 - 0.12x\). At \(x = 0\), \(R''(0) = 96 > 0\) so \(x = 0\) is a local minimum and at \(x = 1600\), \(R''(1600) = 96 - 0.12*1600 = -96 < 0\) so \(x = 1600\) is a local maximum. Considering \(R(0) = 48*0 - 0.02*0 = 0\) and \(R(1600) = 48*1600^2 - 0.02*1600^3\) clearly \(R(1600) > R(0)\). Therefore \(x = 1600\) yields the maximum revenue.

Key concepts

Derivative Calculus

Derivative calculus provides a method to find the rate at which one quantity changes concerning another. This is particularly useful in scenarios where you need to find the maximum or minimum value of a function, such as maximizing revenue. The process involves finding the derivative of a given function, which is essentially a new function representing the slope of the tangent line at any point along the original function.

The derivative provides critical insights into how the original function behaves, indicating intervals where the function is increasing, decreasing, or where it might experience significant changes. By calculating and examining the derivative, we can identify critical points that help determine the maximum or minimum values of the function, paving the way for optimized decisions.

Polynomial Functions

Polynomial functions are algebraic expressions that involve sums of powers of a variable multiplied by coefficients. They are written in the form, for example, of \( ax^n + bx^{n-1} + cx^{n-2} + ... \). These functions are quite common and include linear, quadratic, cubic, and higher-degree polynomials.

Polynomial functions have several characteristic features worth noting:

• They are continuous and smooth, meaning they have no breaks or sharp turns.
• The highest power of the variable indicates the degree of the polynomial, which helps in understanding its graph's general shape.
• They can model real-world scenarios, such as calculating revenues or costs.

Understanding their behavior is crucial when applying techniques like differentiation to find maximum or minimum values.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are significant because they indicate potential maximum or minimum values of the original function, which are particularly important in applications like revenue maximization. To find critical points, you first calculate the derivative of the function and then solve the equation where this derivative equals zero.

The reasoning is simple: if the derivative is zero, the slope of the tangent line is horizontal at that point, which could mean a peak or valley in the function. Once calculated, critical points require further analysis to determine their nature—whether they correspond to a local maximum, local minimum, or just a point of inflection.

Second Derivative Test

The second derivative test is a valuable method for classifying critical points of a function as either maxima, minima, or points of inflection. After identifying critical points, the next step is to compute the second derivative of the function.

The principle behind this test is straightforward:

• If the second derivative at a critical point is positive, the function is concave up at that point, indicating a local minimum.
• Conversely, if the second derivative is negative, the function is concave down, indicating a local maximum.
• If the second derivative is zero, the test is inconclusive, and other methods must be employed.

This technique is instrumental in determining the optimal solution as seen in our revenue maximization exercise. By applying the second derivative test, one can confidently assert whether the critical points found indeed represent maximum revenue points, ensuring that the calculation leads to practical, actionable insights.