Question

Find the number of units \(x\) that produces a maximum revenue \(R\). $$ R=30 x^{2 / 3}-2 x $$

Short answer

The number of units 'x' that produces maximum revenue 'R' is 1000 units.

Step-by-step solution

Write Down the Revenue Function

The given revenue function is \( R=30 x^{2/3} - 2x \)

Derive the Revenue Function

Differentiate the revenue function to find out R'(x). The derivative of \( R(x)=30 x^{2/3} - 2x \) using the power rule is \( R'(x) = 20x^{-1/3} - 2 \)

Find Maximum Revenue

Set the derivative (R'(x)) to zero and solve for x to find the number of units that results in maximum revenue. So, \(20x^{-1/3} - 2 = 0 \Rightarrow x^{1/3} = 10 \Rightarrow x = 1000\)

Verify the Solution

Verify the solution x=1000 as the number of units that give maximum revenue by using the second derivative test. Calculate the second derivative of R(x), R''(x) = -20/3*x^{-4/3}. Insert the solution x = 1000 into R''(x). If the result is negative, x=1000 would be a maximum point as expected. \( R''(1000) = -20/3*1000^{-4/3} < 0 \), so indeed, x=1000 is a maximum point.

Key concepts

Revenue Function

A revenue function, symbolically represented as R(x), is a mathematical model that expresses the total revenue generated by the sale of x units of a product or service. For many businesses, understanding and modeling this function is critical for making informed decisions about pricing and production.

In the given exercise, the revenue function is defined as R=30x^{2/3}-2x. This particular function indicates that the revenue generated is a result of two contrasting terms: the first positive term which increases with the number of units sold, and the second negative term representing the costs that scale linearly with the amount of units. This kind of function often arises in economics where diminishing returns are present, common in production and sales scenarios.

Derivative Application

Applying derivative in calculus is one of the most powerful tools used for optimization problems, such as identifying maximum or minimum values of functions. The first derivative of a function, noted as R'(x), shows the rate at which the function's value is changing at any given point.

In economic terms, it signifies the additional revenue that would be generated by selling one more unit, or conversely, the lost revenue if sales decrease by one unit. By setting this first derivative equal to zero, we find critical points where the rate of change is neither increasing nor decreasing; these critical points are possible candidates for maximum or minimum revenue.

Power Rule in Differentiation

One of the most fundamental rules in calculus for finding derivatives is the power rule. It states that if you have a function f(x) = ax^n, where a is a constant and n is a real number, the derivative of f(x) with respect to x is f'(x) = anx^{n-1}.

When applying the power rule to the revenue function provided in the exercise, we differentiate each term separately. The differentiation of 30x^{2/3} following the power rule gives us 20x^{-1/3}, while the differentiation of the linear term -2x simply yields -2 because the power of x is 1, and subtracting 1 from the exponent results in x^0, which is 1.

Second Derivative Test

The second derivative test is a technique used to determine whether a given critical point of a function is a local maximum or minimum. If the second derivative at a critical point is negative (R''(x)<0), the function has a local maximum at that point. Conversely, if the second derivative is positive (R''(x)>0), the function has a local minimum.

In the context of the problem we're discussing, once we've found a critical point by setting the first derivative to zero, we then take the second derivative of the revenue function. Upon applying the second derivative test to our critical point, we find that the second derivative is negative, indicating a local maximum revenue at that level of production. This confirms that the business would maximize its revenue by selling 1000 units.