Suppose that the total revenue function for a manufacturer...

Suppose that the total revenue function for a manufacturer is $R(x)=300 \ln (x+1)$, so the sale of $x$ units of a product brings in about $R(x)$ dollars. Suppose also that the total cost of producing $x$ units is $C(x)$ dollars, where $C(x)=2 x .$ Find the value of $x$ at which the profit function $R(x)-C(x)$ will be maximized. Show that the profit function has a relative maximum and not a relative minimum point at this value of $x$.

Key Concepts

Critical Points Determination

Finding critical points, where the derivative of the profit function is zero or undefined, is a key step. This process identifies values of the independent variable that can potentially maximize or minimize the profit. This technique is central to many optimization problems across economics and engineering.

Logarithmic Functions in Revenue Models

Logarithmic functions often appear in revenue models to reflect scenarios where growth decelerates over time or exhibits diminishing returns. Understanding properties of logarithmic functions, such as their rates of change, is essential when modeling revenue behaviors that are not linear.

Second Derivative Test

The second derivative test is used to determine the nature of a critical point once it has been found. By evaluating the second derivative at the critical point, one can establish whether the function is concave up (indicating a local minimum) or concave down (indicating a local maximum). This test is crucial in confirming the optimal conditions in maximization problems.

Profit Function

A profit function represents the financial gain of a business by subtracting its total costs from its total revenues. In an optimization context, it is used to determine the level of production that maximizes profit by analyzing the difference between how much money is brought in and how much is spent to produce goods or services.

Optimization via Derivatives

Using calculus to optimize functions involves differentiating the profit function with respect to the quantity produced (or another variable) and determining where this derivative equals zero. These critical points are potential candidates for local extrema, wherein the function might reach a maximum or minimum value.

Transcript

00:01 This question we are given that.

00:01 Suppose the total revenue function for the manufacturer is 300 times national log of x plus one.

00:08 So the sale of x units of the product brings in about rx dollars.

00:14 Suppose that the total cost function for x units is given as 2x.

00:19 So we have to find the value of its for which the profit function, which is r minus c is maximized.

00:24 So first let's find the profit function.

00:26 That's nothing but rx minus c is already given.

00:28 So expression for rx is 300 times natural log of x plus 1 and the expression for cost function is 2x.

00:37 Okay, so in order to maximize, definitely we've got to use the derivative of the profit function and equate to 0 to find a critical points and then to find relative maxima as well.

00:48 So p -x is going to be 300 is a constant comes outside.

00:53 Differencesation of natural log of x plus 1.

00:56 So the differentiation of natural log of x is 1 over x plus 1.

00:59 Then we go inside and we differentiate x which is one and then the differentiation of a constant is zero minus two as a constant comes out and the differentiation of x is just one.

01:10 300 over x plus 1 minus 2 as the profit function.

01:16 So in order to find a critical points, we'll equate the profit function to 0.

01:21 So p -dash x is equated to 0...

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