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Let's address the given problem step by step.

### Part a: Profit Function and Maximizing Quantity

1. **Revenue**: The price received per ton is given by \( P = 1000 - \frac{1}{3}Q \), and the quantity sold is \( Q \). Therefore, the total revenue \( R(Q) \) is:

   \[
   R(Q) = Q \times P = Q \times \left(1000 - \frac{1}{3}Q\right)
   \]
   Simplifying:

   \[
   R(Q) = 1000Q - \frac{1}{3}Q^2
   \]

2. **Cost**: The cost per ton consists of two components:
   - The price paid per ton is \( P = 800 + \frac{1}{5}Q \).
   - There is also an additional transportation cost of 100 per ton.

   So, the total cost per ton is:

   \[
   C(Q) = \left(800 + \frac{1}{5}Q\right) + 100 = 900 + \frac{1}{5}Q
   \]

   The total cost for producing \( Q \) tons is:

   \[
   TC(Q) = Q \times (900 + \frac{1}{5}Q) = 900Q + \frac{1}{5}Q^2
   \]

3. **Profit Function**: The profit function \( \pi(Q) \) is the difference between total revenue and total cost:

   \[
   \pi(Q) = R(Q) - TC(Q)
   \]

   Substituting the expressions for \( R(Q) \) and \( TC(Q) \):

   \[
   \pi(Q) = \left(1000Q - \frac{1}{3}Q^2\right) - \left(900Q + \frac{1}{5}Q^2\right)
   \]

   Simplifying:

   \[
   \pi(Q) = 100Q - \left(\frac{1}{3}Q^2 + \frac{1}{5}Q^2\right)
   \]

   Combine the terms involving \( Q^2 \):

   \[
   \pi(Q) = 100Q - \left(\frac{5}{15}Q^2 + \frac{3}{15}Q^2\right)
   \]
   \[
   \pi(Q) = 100Q - \frac{8}{15}Q^2
   \]

4. **Profit Maximization**: To maximize profit, take the derivative of \( \pi(Q) \) with respect to \( Q \) and set it equal to zero:

   \[
   \frac{d\pi}{dQ} = 100 - \frac{16}{15}Q = 0
   \]

   Solving for \( Q \):

   \[
   100 = \frac{16}{15}Q
   \]

   \[
   Q = \frac{100 \times 15}{16} = 93.75
   \]

   Therefore, the profit-maximizing quantity is \( Q = 93.75 \) tons.

---

### Part b: Impact of a Government Tax

If the government imposes a tax of 10 per ton, the total cost will increase by \( 10Q \). The new total cost function becomes:

\[
TC_{\text{new}}(Q) = 900Q + \frac{1}{5}Q^2 + 10Q = 910Q + \frac{1}{5}Q^2
\]

The new profit function \( \hat{\pi}(Q) \) is:

\[
\hat{\pi}(Q) = R(Q) - TC_{\text{new}}(Q)
\]

Substitute the expressions for \( R(Q) \) and \( TC_{\text{new}}(Q) \):

\[
\hat{\pi}(Q) = \left(1000Q - \frac{1}{3}Q^2\right) - \left(910Q + \frac{1}{5}Q^2\right)
\]

Simplifying:

\[
\hat{\pi}(Q) = 90Q - \left(\frac{1}{3}Q^2 + \frac{1}{5}Q^2\right)
\]

\[
\hat{\pi}(Q) = 90Q - \frac{8}{15}Q^2
\]

To maximize the new profit, take the derivative of \( \hat{\pi}(Q) \) with respect to \( Q \):

\[
\frac{d\hat{\pi}}{dQ} = 90 - \frac{16}{15}Q = 0
\]

Solve for \( Q \):

\[
90 = \frac{16}{15}Q
\]

\[
Q = \frac{90 \times 15}{16} = 84.375
\]

Therefore, with the tax, the new profit-maximizing quantity is \( Q = 84.375 \) tons.

---

Would you like further details or explanations on any part of the solution?

### Related Questions:
1. How would the result change if the transportation cost per ton were higher?
2. What if the price paid per ton had a different slope in its function?
3. How does a change in the tax rate affect the firm's profit-maximizing quantity?
4. What would happen if the price received per ton were fixed instead of dependent on \( Q \)?
5. How would a subsidy instead of a tax influence the profit-maximizing quantity?

### Tip:
When maximizing profit, always check the second derivative to confirm whether you have a maximum or minimum.

If a firm sells Q tons of a product, the price P received per ton is P = 1000 - (1/3)Q. The price it has to pay per ton is P = 800 + (1/5)Q. In addition, it has transportation costs of 100 per ton. a) Express the firm's profit as a function of Q, the number of tons sold, and find the profit-maximizing quantity. b) Suppose the government imposes a tax on the firm's product of 10 per ton. Find the new expression for the firm's profits and the new profit-maximizing quantity.

Explore how to maximize a firm's profit using revenue and cost functions with quadratic terms. Learn how government taxes affect the profit-maximizing quantity. The problem covers calculus-based optimization techniques.

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