Marginal Analysis

Marginal Cost & Marginal Average Cost Functions

Definition

If \(C(x)\) denotes the total cost function, then \(C'(x)\) denotes the marginal cost function, which approximates the extra cost incurred in producing one additional unit

\[C'(x) \approx C(x+1) - C(x)\]

and \(\overline{C}'(x)\) denotes the marginal average cost function.

Marginal Revenue & Marginal Average Revenue Functions

Definition

If \(R(x)\) denotes the total revenue function, then \(R'(x)\) denotes the marginal revenue function, which approximates the revenue realized from the sale of one additional unit

\[R'(x) \approx R(x+1) - R(x)\]

and \(\overline{R}'(x)\) denotes the marginal average revenue function.

Marginal Profit & Marginal Average Profit Functions

Definition

If \(P(x)\) denotes the total profit function, then \(P'(x)\) denotes the marginal profit function, which approximates the profit generated from the production and sale of one additional unit

\[P'(x) \approx P(x+1) - P(x)\]

and \(\overline{P}'(x)\) denotes the marginal average profit function.

Marginal Analysis Notation

Name	Function	Average	Marginal Function	Marginal Average
Cost	\(C(x)\)	\(\overline{C}(x) = \frac{C(x)}{x}\)	\(C'(x)\)	\( \overline{C}'(x) = \frac{d}{dx}\overline{C}(x)\)
Revenue	\(R(x)\)	\(\overline{R}(x) = \frac{R(x)}{x}\)	\(R'(x)\)	\( \overline{R}'(x) = \frac{d}{dx}\overline{R}(x)\)
Profit	\(P(x)\)	\(\overline{P}(x) = \frac{P(x)}{x}\)	\(P'(x)\)	\(\overline{P}'(x) = \frac{d}{dx}\overline{P}(x)\)

Example 1

The daily demand for the new PBox5 Game Dr. Mathematica-Exam Day of Reckoning is given by

\[p(x) = \frac{125}{x+2} - \frac{1}{2}x\]

where \(x\) is the number of video games sold each day and \(p\) is in dollars. Using the marginal revenue function, \(R'(x)\), approximate the marginal revenue when 3 video games are sold each day and interpret the result.

Step 1:   Compute the total revenue function,   \(R(x)\).

\[\begin{align*} R(x) &= x\cdot p(x) \\ &= x\left(\frac{125}{x+2} - \frac{1}{2}x \right)\\ &= \frac{125x}{x+2}-\frac{1}{2}x^2 \end{align*}\]

Step 2:   Compute the marginal revenue function,   \(R'(x)\).

\[\begin{align*} R'(x) &= \frac{d}{dx}R(x)\\ &= \frac{d}{dx}\left( \frac{125x}{x+2}-\frac{1}{2}x^2 \right) \\ &= \frac{125(x+2) -125x}{(x+2)^2} - \frac{1}{2}\cdot 2x \\ &= \frac{125x+250 - 125x}{(x+2)^2} - x \\ &= \frac{250}{(x+2)^2} - x \\ \end{align*}\]

Step 3:   Plug in   \(x=3\)   into the marginal revenue function.

\[\begin{align*} R'(3) &= \frac{250}{(3+2)^2} - 3 \\ &= \frac{250}{5^2} - 3 \\ %&= \frac{250}{25} - 3 \\ &= 10 - 3\\ &= 7 \end{align*}\]

Therefore, the total daily revenue would increase by approximately $7 if sales increased from 3 to 4 units each day.

Example 2

If the demand function for math self-help videos is given by

\[p(x) = 35 - 0.1x\]

and the total cost function to manufacture the videos is given by

\[C(x) = 3x + 21\]

Evaluate the marginal profit function at \(x=20\) and interpret the result.

Step 1:   Compute the total revenue function,   \(R(x)\).

\[\begin{align*} R(x) &= x\cdot p(x)\\ &= x(35 - 0.1x)\\ &= 35x-0.1x^2 \end{align*}\]

Step 2:   Compute the total profit function,   \(P(x)\).

\[\begin{align*} P(x) &= R(x) - C(x)\\ &= (35x-0.1x^2)-(3x+21)\\ &= 35x-0.1x^2-3x-21\\ &=-0.1x^2 +32x -21 \end{align*}\]

Step 3:   Compute the marginal profit function,   \(P'(x)\).

\[\begin{align*} P'(x) &= \frac{d}{dx}P(x) \\ &= \frac{d}{dx}(-0.1x^2 +32x -21) \\ &= -0.1(2x) + 32 \\ &= -0.2x+32 \end{align*}\]

Step 4:   Plug in   \(x=20\)   into the marginal profit function.

\[\begin{align*} P'(20) &= -0.2(20)+32\\ &= -\frac{2}{10}(20)+32\\ &=-4+32\\ &=28 \end{align*}\]

Therefore, the total profit will increase by approximately $28 when the 21st video is produced and sold.

Example 3

The daily cost (in dollars) of producing computer screens is given by

\[C(x)=9x^3-30x^2+90x+900\]

where \(x\) denotes the number of thousands of screens produced each day. Calculate the marginal average cost function when 3000 screens are produced each day and interpret the result.

Step 1:   Compute and simplify the average cost function,   \(\overline{C}(x)\).

\[\begin{align*} \overline{C}(x) &= \frac{9x^3-30x^2+90x+900}{x} && \hbox{since $\overline{C}(x) = C(x)/x$}\\ \\ &= \frac{9x^3}{x} -\frac{30x^2}{x}+\frac{90x}{x}+\frac{900}{x}&& \text{separate the terms to make differentiation easier}\\ \\ &=9x^2 -30x+90+900x^{-1} && \text{simplify}\\ \end{align*}\]

Step 2:   Compute the marginal average cost function,   \(\overline{C}'(x)\).

\[\begin{align*} \overline{C}'(x) &= \frac{d}{dx} \overline{C}(x)\\ &= \frac{d}{dx}\left(9x^2 -30x+90+900x^{-1}\right)\\ &= 9(2x)-30+ 0 + (-1)900x^{-2}\\ &=18x-30-\frac{900}{x^2} \end{align*}\]

Step 3:   Plug in   \(x=3\).

We plug in \(x=3\), since \(x\) denotes the number of thousands of screens produced each day.

\[\begin{align*} \overline{C}'(3) &=18(3)-30-\frac{900}{3^2} \\ &=54-30-100 \\ &= -76 \end{align*}\]

Therefore, the average cost of producing a thousand screens will decrease by approximately $76 if the level of production is increased from 3000 to 4000 screens.