Profit Maximization Output Calculation

To find the output level ($q$) that maximizes profit, we need to set up the profit function and find its maximum point using calculus.

1. Determine the Profit Function

• Profit ($\pi$) equals Total Revenue (TR) minus Total Cost (TC).
• The Total Cost function is given as: $C = q^3 - 10q^2 + 17q + 66$.
• The Price ($p$) is fixed at $5$.
• Total Revenue (TR) is calculated as $p \times q$. Therefore, $TR = 5q$.
• The profit function is:

  $\pi(q) = TR - C$

  $\pi(q) = 5q - (q^3 - 10q^2 + 17q + 66)$

  $\pi(q) = -q^3 + 10q^2 - 12q - 66$.

2. Apply Calculus for Profit Maximization

• Profit is maximized when the first derivative of the profit function with respect to quantity ($q$) is zero ($\frac{d\pi}{dq} = 0$). This is the first-order condition (FOC).
• Calculate the first derivative:

  $\frac{d\pi}{dq} = \frac{d}{dq}(-q^3 + 10q^2 - 12q - 66)$

  $\frac{d\pi}{dq} = -3q^2 + 20q - 12$.

• Set the first derivative to zero to find critical points:

  $-3q^2 + 20q - 12 = 0$.

• To simplify, multiply the equation by -1:

  $3q^2 - 20q + 12 = 0$.

3. Solve for Output Quantity ($q$)

• Use the quadratic formula $q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=3$, $b=-20$, and $c=12$.
• $q = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(3)(12)}}{2(3)}$
• $q = \frac{20 \pm \sqrt{400 - 144}}{6}$
• $q = \frac{20 \pm \sqrt{256}}{6}$
• $q = \frac{20 \pm 16}{6}$.
• This yields two possible values for $q$:

  $q_1 = \frac{20 + 16}{6} = \frac{36}{6} = 6$.

  $q_2 = \frac{20 - 16}{6} = \frac{4}{6} = \frac{2}{3}$.

4. Verify the Profit Maximization Point

• Check the second-order condition (SOC): the second derivative must be negative ($\frac{d^2\pi}{dq^2} < 0$) for a maximum.
• Calculate the second derivative:

  $\frac{d^2\pi}{dq^2} = \frac{d}{dq}(-3q^2 + 20q - 12)$

  $\frac{d^2\pi}{dq^2} = -6q + 20$.

• Evaluate the SOC at the critical points:
  • For $q=6$: $\frac{d^2\pi}{dq^2} = -6(6) + 20 = -36 + 20 = -16$. Since $-16 < 0$, profit is maximized at $q=6$.
  • For $q=2/3$: $\frac{d^2\pi}{dq^2} = -6(2/3) + 20 = -4 + 20 = 16$. Since $16 > 0$, profit is minimized at $q=2/3$.

5. Final Answer

• The output level at which profit is maximized is $q=6$.