Question

If
C(x) = 18,000 + 400x − 0.6x2 + 0.004x3
is the cost function and
p(x) = 1,600 − 6x
is the demand function, find the production level (in units) that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.)

Answer 1

To find the production level that maximizes profit, we first need to express profit in terms of quantity \( x \). Profit \( \Pi(x) \) is given by the difference between total revenue \( R(x) \) and total cost \( C(x) \):

\[
\Pi(x) = R(x) - C(x)
\]

### Step 1: Find Total Revenue \( R(x) \)

Total revenue is calculated as the product of price \( p(x) \) and quantity \( x \):

\[
R(x) = p(x) \cdot x = (1,600 - 6x) \cdot x
\]

Expanding this:

\[
R(x) = 1,600x - 6x^2
\]

### Step 2: Write the Profit Function \( \Pi(x) \)

Now substituting the revenue and cost functions, we have:

\[
\Pi(x) = R(x) - C(x) = (1,600x - 6x^2) - (18,000 + 400x - 0.6x^2 + 0.004x^3)
\]

Simplifying this:

\[
\Pi(x) = 1,600x - 6x^2 - 18,000 - 400x + 0.6x^2 - 0.004x^3
\]

Combining like terms:

\[
\Pi(x) = (1,600x - 400x) + (-6x^2 + 0.6x^2) - 0.004x^3 - 18,000
\]
\[
\Pi(x) = 1,200x - 5.4x^2 - 0.004x^3 - 18,000
\]

### Step 3: Marginal Revenue and Marginal Cost

To find the production level that maximizes profit, we need to take the derivative of the revenue and cost functions.

**Marginal Revenue \( MR \)** is the derivative of \( R(x) \):

\[
MR = \frac{dR}{dx} = 1,600 - 12x
\]

**Marginal Cost \( MC \)** is the derivative of \( C(x) \):

\[
MC = \frac{dC}{dx} = 400 - 1.2x + 0.012x^2
\]

### Step 4: Set Marginal Revenue Equal to Marginal Cost

To find the profit-maximizing quantity, set \( MR = MC \):

\[
1,600 - 12x = 400 - 1.2x + 0.012x^2
\]

Reorganizing this equation:

\[
0.012x^2 + 10.8x - 1,200 = 0
\]

### Step 5: Solve the Quadratic Equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 0.012 \), \( b = 10.8 \), and \( c = -1,200 \).

Calculate the discriminant:

\[
D = b^2 - 4ac = (10.8)^2 - 4 \cdot 0.012 \cdot (-1,200)
\]
\[
D = 116.64 + 57.6 = 174.24
\]

Now apply the quadratic formula:

\[
x = \frac{-10.8 \pm \sqrt{174.24}}{2 \cdot 0.012}
\]
Calculate \( \sqrt{174.24} \approx 13.2 \):

\[
x = \frac{-10.8 \pm 13.2}{0.024}
\]

This gives us two potential solutions:

1. \( x = \frac{2.4}{0.024} = 100 \)
2. \( x = \frac{-24}{0.024} = -1000 \) (not a feasible solution)

Thus, the production level that maximizes profit is:

\[
\boxed{100}
\] units.