Questions: MTH 263 Homework Ch 2.7 - MIH 26 The cost (in dollars) of producing x units of a certain commodity is C(x)=5,000 (a) Find the average rate of change (in per unit) of C with respect to x when nearest cent.) (i) x=103 26 per unit (ii) x=101 32.1 per unit (b) Find the instantaneous rate of change (in per unit) of C with respect to x chapter.) 32.0 per unit

Solution Steps

To solve this problem, we need to:

1. Calculate the average rate of change of the cost function \( C(x) \) with respect to \( x \) for the given values of \( x \). The average rate of change is given by the formula \(\frac{C(x_2) - C(x_1)}{x_2 - x_1}\).
2. Find the instantaneous rate of change of the cost function \( C(x) \) with respect to \( x \). This is the derivative of \( C(x) \).

To find the average rate of change of the cost function \( C(x) \) when \( x = 103 \), we use the formula:

\[ \text{Average Rate of Change} = \frac{C(x_2) - C(x_1)}{x_2 - x_1} \]

Substituting \( C(x) = 5000 \), \( x_1 = 100 \), and \( x_2 = 103 \):

\[ \text{Average Rate of Change} = \frac{5000 - 5000}{103 - 100} = \frac{0}{3} = 0.0 \]

Next, we calculate the average rate of change when \( x = 101 \):

\[ \text{Average Rate of Change} = \frac{C(x_2) - C(x_1)}{x_2 - x_1} \]

Substituting \( x_2 = 101 \):

\[ \text{Average Rate of Change} = \frac{5000 - 5000}{101 - 100} = \frac{0}{1} = 0.0 \]

The instantaneous rate of change of the cost function \( C(x) \) is given by the derivative of \( C(x) \). Since \( C(x) = 5000 \) is a constant function, its derivative is:

\[ \frac{dC}{dx} = 0 \]

Final Answer