Cost and Revenue

Let \(d = C(x)\) represent the total dollar amount spent on producing \(x\) items. The average rate of change in cost from the first item produced \(x_1\) and the last item produced \(x_2\) is

\[ \begin{align} \mbox{Average Rate of Change of Cost } &= \frac{\Delta d}{\Delta x} \\ &= \frac{C(x_2) - C(x_1)}{x_2-x_1}. \end{align}\]

Similarly, let \(p = R(x)\) represent the total dollar amount made from selling \(x\) items. The average rate of change in revenue from the first item sold \(x_1\) and the last item sold \(x_2\) is

\[ \begin{align} \mbox{Average Rate of Change in Revenue } &= \frac{\Delta p}{\Delta x} \\ &= \frac{R(x_2) - R(x_1)}{x_2-x_1}.\end{align} \]

Again, let \(d = C(x)\) represent the total dollar amount spent on producing \(x\) items. The instantaneous rate of change of cost when producing item \(x\) is

\[ \begin{align} \mbox{Instantaneous Rate of Change of Cost } &= \lim\limits_{\Delta x \to 0} \frac{\Delta d}{\Delta x} \\ &= \frac{dd}{dx}. \end{align}\]

Similarly, let \(p = R(x)\) represent the total dollar amount made from selling \(x\) items. The instantaneous rate of change in revenue when selling item \(x\) is

\[ \begin{align} \mbox{Rate of Change in Revenue } &= \lim\limits_{\Delta x \to 0} \frac{\Delta p}{\Delta x} \\ &= \frac{dp}{dx}. \end{align}\]

A newspaper company has a fixed cost of production of \(\$80\) per edition, as well as a marginal distribution and printing materials cost of \(40¢\) per copy. Newspapers are sold at \(50¢\) per copy.

1. Formulate functions for the cost, revenue, and profit of the newspaper.
2. Then, find out whether the newspaper makes a profit or not from selling \(600\) copies.
3. How many copies of the newspaper have to be sold to break even?

Solutions:

1. Formulate functions for the cost, revenue, and profit of the newspaper.
   • As mentioned in the Cost and Revenue Formula section above, you need to use the context of the given problem to determine the formulas for cost, revenue, and profit.
     1. Based on the costs that the problem provides you, the cost function can be written as the fixed cost plus the cost per copy.\[C(x) = 80 + 0.4x\]
     2. Similarly, the revenue equation can be written as the amount the newspaper company earns per copy.\[R(x) = 0.5x\]
     3. Finally, since profit is the company's revenue minus its cost:\[ \begin{align}P(x) &= R(x) - C(x) \\&= 0.5x - (0.4x + 80) \\P(x) &= 0.1x-80\end{align}\]
2. Find out whether the newspaper makes a profit or not from selling \(600\) copies.
   1. Plug \(x = 600\) into your profit equation.\[P(600) = 0.1(600)-80 = -20\]
   2. Since the profit is negative, the newspaper company does not make a profit from selling \(600\) copies, but rather incurs a loss.
3. Find out how many copies of the newspaper have to be sold to break even.
   • Breaking even means that the company does not earn any profit. So, the cost of production and the earned revenue are equal.
   • To find out how many newspapers the company must sell to break even, you need to set the profit equation equal to \(0\), or \(P(x) = 0\) and then solve for \(x\).
     1. Set \(P(x) = 0\).\[ P(x) = 0.1x - 80 = 0 \]
     2. Solve for \(x\).\[ \begin{align}0.1x - 80 &= 0 \\0.1x &= 80 \\x &= \frac{80}{0.1} \\x &= 800\end{align} \]
   • Therefore, the newspaper must sell \(800\) copies to break even.

Suppose a toy company produces \(x\) toys with a cost of \(C(x) = 10000 + 3x + 0.01x^2\) that encompasses the total overhead costs (like rent for the factory), materials, labor, and so on.

1. Find the marginal cost of producing \(500\) toys.
2. What does the company have to charge for the \(500^{th}\) toy to make a profit of \(\$10\)?
3. Then, find the average marginal cost from \(200\) toys to \(300\) toys.

Solutions:

1. Find the marginal cost of producing \(500\) toys.
   1. Find the marginal cost equation.
      • Remember, the marginal cost equation is the derivative of the cost equation. So, use the power rule to take the derivative of \(C(x)\).\[ C'(x) = 3 +0.02x\]
   2. Plug in the number of toys for \(x\) to find the cost of producing a toy.
      • Plugging in \(x = 500\) to \(C(x)\) gives you\[ C(500) = 3 + 0.02(500) = 13.\]
      • What does this value actually mean?\(C(500) = 13\) means that when producing the \(500^{th}\) toy, the company is building toys at a cost of \(\$13\) per toy.
      • Therefore, the marginal cost of producing the \(500\) toys is \(\$13\).
2. What does the company have to charge for the \(500^{th}\) toy to make a profit of \($10\)?
   • From solution A, you know that the \(500^{th}\) toy cost \(\$13\) for the company to build. And, since you know the equation for profit is \(P(x) = R(x) - C(x)\), you can solve this equation for revenue to determine how much the company should charge for the toy.
     1. Use the equation for profit:\[ P(x) = R(x) - C(x) \]
     2. Solve for revenue.\[ \begin{align}P(x) &= R(x) - C(x) \\P(x) + C(x) &= R(x) \\R(x) &= P(x) + C(x)\end{align} \]
     3. Plug in your known values for profit \(($10)\) and cost \(($13)\), and solve for revenue.\[ R(x) = \$10 + \$13 = \$23 \]
   • In order for the company to make a profit of \(\$10\) for the \(500^{th}\) toy built, they must charge \(\$23\) for it.
3. Find the average marginal cost from \(200\) toys to \(300\) toys.
   1. To find the average marginal cost from \(200\) toys to \(300\) toys, you need to use the average rate of change formula:\[ \mbox{Average Rate of Change } = \frac{\Delta d}{\Delta x} = \frac{C(x_2) - C(x_1)}{x_2-x_1}\]where,\[ x_1 = 200 \text{ and } x_2 = 300. \]
   2. Next, you need to use the cost function given to find \(C(x_1)\) and \(C(x_2)\).
      • For \(x_1 = 200\):\[ \begin{align}C(x_1) &= 10000 + 3(x_1) + 0.01(x_1)^2 \\C(200) &= 10000 + 3(200) + 0.01(200)^2 \\C(200) &= 10000 + 600 + 400 \\C(200) &= 11000\end{align} \]
      • For \(x_2 = 300\):\[ \begin{align}C(x_2) &= 10000 + 3(x_2) + 0.01(x_2)^2 \\C(300) &= 10000 + 3(300) + 0.01(300)^2 \\C(300) &= 10000 + 900 + 900 \\C(300) &= 11800\end{align} \]
   3. Now, plug all your values into the average rate of change formula.\[ \mbox{Average Rate of Change } = \frac{\Delta d}{\Delta x} = \frac{C(x_2) - C(x_1)}{x_2-x_1}\]where,\[ x_1 = 200, x_2 = 300, C(x_1) = 11000, \text{ and } C(x_2) = 11800. \]\[ \begin{align}\mbox{Average Rate of Change } &= \frac{\Delta d}{\Delta x} \\&= \frac{C(x_2) - C(x_1)}{x_2-x_1} \\&= \frac{11800 - 11000}{300 - 200} \\&= \frac{800}{100} \\&= 8\end{align} \]
   4. So, on average, the cost of producing a toy between the \(200^{th}\) toy built and the \(300^{th}\) toy built is \(\$8\) per toy.