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:
- \( x = \frac{2.4}{0.024} = 100 \)
- \( x = \frac{-24}{0.024} = -1000 \) (not a feasible solution)
Thus, the production level that maximizes profit is:
\[ \boxed{100} \] units.
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