Understanding Total Cost and Average Cost

The question asks us to find the minimum average cost (AC) and the level of output (Q) at which this minimum cost occurs, given the total cost (TC) function: \( TC = Q^3 - 10Q^2 + 60Q \).

In economics, the average cost is calculated by dividing the total cost (TC) by the quantity of output (Q).

The formula for Average Cost (AC) is:

\[ AC = \frac{TC}{Q} \]

Given the total cost function \( TC = Q^3 - 10Q^2 + 60Q \), we can find the average cost function by dividing each term by Q (assuming \( Q > 0 \)):

\[ AC = \frac{Q^3 - 10Q^2 + 60Q}{Q} = \frac{Q^3}{Q} - \frac{10Q^2}{Q} + \frac{60Q}{Q} \]

This simplifies to:

\[ AC = Q^2 - 10Q + 60 \]

Finding the Level of Output for Minimum Average Cost

To find the minimum point of a cost function, we typically use calculus. The minimum (or maximum) of a function occurs where its first derivative is equal to zero. So, we need to find the derivative of the average cost function with respect to Q and set it to zero.

The average cost function is \( AC(Q) = Q^2 - 10Q + 60 \). Let's find the first derivative \( \frac{d(AC)}{dQ} \):

\[ \frac{d(AC)}{dQ} = \frac{d}{dQ}(Q^2 - 10Q + 60) \]

\[ \frac{d(AC)}{dQ} = 2Q - 10 \]

Now, set the first derivative equal to zero to find the critical point(s):

\[ 2Q - 10 = 0 \]

\[ 2Q = 10 \]

\[ Q = \frac{10}{2} \]

\[ Q = 5 \]

So, a critical point for the average cost occurs at an output level of \( Q = 5 \).

Confirming Minimum Average Cost using the Second Derivative Test

To confirm whether this output level \( Q = 5 \) corresponds to a minimum average cost, we can use the second derivative test. We need to find the second derivative of the average cost function and evaluate it at \( Q = 5 \).

The first derivative was \( \frac{d(AC)}{dQ} = 2Q - 10 \). Let's find the second derivative \( \frac{d^2(AC)}{dQ^2} \):

\[ \frac{d^2(AC)}{dQ^2} = \frac{d}{dQ}(2Q - 10) \]

\[ \frac{d^2(AC)}{dQ^2} = 2 \]

Since the second derivative \( \frac{d^2(AC)}{dQ^2} = 2 \) is positive (\( > 0 \)) for all values of Q (including \( Q = 5 \)), the average cost function has a minimum at \( Q = 5 \).

Calculating the Minimum Average Cost

Now that we know the minimum average cost occurs at an output level of \( Q = 5 \), we can substitute this value back into the average cost function to find the minimum average cost itself.

The average cost function is \( AC(Q) = Q^2 - 10Q + 60 \). Substitute \( Q = 5 \):

\[ AC_{min} = (5)^2 - 10(5) + 60 \]

\[ AC_{min} = 25 - 50 + 60 \]

\[ AC_{min} = -25 + 60 \]

\[ AC_{min} = 35 \]

Therefore, the minimum average cost is 35, and it occurs at an output level of \( Q = 5 \).

Summary of Results

Based on our calculations:

• The level of output at which minimum average cost occurs is \( Q = 5 \).
• The minimum average cost at this output level is \( AC_{min} = 35 \).

Revision Table: Key Cost Concepts

Additional Information: Minimum Average Cost and Marginal Cost

An important concept in cost theory is the relationship between average cost (AC) and marginal cost (MC). The marginal cost is the derivative of the total cost function with respect to output Q:

\[ TC = Q^3 - 10Q^2 + 60Q \]

\[ MC = \frac{d(TC)}{dQ} = \frac{d}{dQ}(Q^3 - 10Q^2 + 60Q) \]

\[ MC = 3Q^2 - 20Q + 60 \]

The average cost curve reaches its minimum point where the marginal cost curve intersects it. This means that at the minimum of the average cost, \( MC = AC \).

Let's check this for our problem:

We found that minimum AC occurs at \( Q = 5 \). At \( Q = 5 \), \( AC = 35 \). Let's calculate MC at \( Q = 5 \):

\[ MC = 3(5)^2 - 20(5) + 60 \]

\[ MC = 3(25) - 100 + 60 \]

\[ MC = 75 - 100 + 60 \]

\[ MC = -25 + 60 \]

\[ MC = 35 \]

Indeed, at \( Q = 5 \), \( MC = 35 \) and \( AC = 35 \). This confirms that the minimum average cost occurs at the point where marginal cost equals average cost.

This relationship is a fundamental principle in microeconomics when analyzing cost functions.