To solve for the quantity $Q$ that maximizes profit, we need to use the given equations for Total Revenue (TR) and Total Cost (TC), then derive the Profit function $\pi(Q)$.

Step 1: Define the profit function

The profit function is defined as: $\pi(Q) = TR - TC$

Given: $TR = 5,000Q - 10Q^2$ $TC = 5Q^2 - 10,000$

Substitute $TR$ and $TC$: $\pi(Q) = (5,000Q - 10Q^2) - (5Q^2 - 10,000)$ $\pi(Q) = 5,000Q - 10Q^2 - 5Q^2 + 10,000$ $\pi(Q) = 5,000Q - 15Q^2 + 10,000$

Step 2: Find the critical point

To find the quantity $Q$ that maximizes profit, take the derivative of $\pi(Q)$ with respect to $Q$, and set it equal to zero: $\frac{d\pi(Q)}{dQ} = 5,000 - 30Q = 0$

Solve for $Q$: $30Q = 5,000$ $Q = \frac{5,000}{30} = 166.67$

Step 3: Determine the maximum profit

Now substitute $Q = 166.67$ back into the profit function to find the maximum profit: $\pi(166.67) = 5,000(166.67) - 15(166.67)^2 + 10,000$ Let's compute this:

$\pi(166.67) = 833,350 - 15 \cdot 27,777.78 + 10,000$ $\pi(166.67) = 833,350 - 416,666.7 + 10,000$ $\pi(166.67) = 426,683.3$

Final Answer

• The quantity $Q$ that maximizes profit is 166.67.
• The maximum profit is approximately 426,683.3.

Would you like a more detailed explanation or have any questions?

5 Related Questions:

1. How do you determine the derivative of a profit function?
2. What is the significance of the second derivative in finding maxima or minima?
3. How do fixed costs affect the profit-maximizing quantity?
4. Can this problem be solved using a graphical approach?
5. How would the profit-maximizing $Q$ change if TC was different?

Tip:

Remember that when maximizing or minimizing a function, the first derivative helps find critical points, while the second derivative test confirms whether it's a maximum or minimum!