The weekly demand function for x units of a product sold by only one firm is 
p = 300 − 
1
2
x dollars,
 and the average cost of production and sale is 
C = 200 + 2x dollars.
(a) Find the quantity that will maximize profit.
 units

(b) Find the selling price at this optimal quantity.
$ 
 per unit

(c) What is the maximum profit?
$ 


To solve this problem, we will calculate the **profit function**, find its maximum, and derive the optimal quantity and price.

### Step 1: Profit Function
The **profit** is defined as:
\[
\text{Profit} = \text{Revenue} - \text{Cost}
\]

- Revenue ($R$): $ R = p \cdot x = (300 - \frac{1}{2}x) \cdot x $
\[
R = 300x - \frac{1}{2}x^2
\]

- Cost ($C$): $C = 200 + 2x$

Thus, the **profit function** becomes:
\[
P(x) = R - C = (300x - \frac{1}{2}x^2) - (200 + 2x)
\]
\[
P(x) = 300x - \frac{1}{2}x^2 - 200 - 2x
\]
\[
P(x) = -\frac{1}{2}x^2 + 298x - 200
\]

### Step 2: Maximizing Profit
To maximize profit, take the derivative of $P(x)$ and set it equal to zero:
\[
P'(x) = -x + 298
\]
\[
-x + 298 = 0
\]
\[
x = 298
\]

Thus, the quantity that maximizes profit is:
\[
\boxed{298 \text{ units}}
\]

### Step 3: Selling Price at Optimal Quantity
The selling price is given by the demand function:
\[
p = 300 - \frac{1}{2}x
\]
Substitute $x = 298$:
\[
p = 300 - \frac{1}{2}(298)
\]
\[
p = 300 - 149 = 151
\]

Thus, the selling price at the optimal quantity is:
\[
\boxed{\$151 \text{ per unit}}
\]

### Step 4: Maximum Profit
Substitute $x = 298$ into the profit function:
\[
P(x) = -\frac{1}{2}x^2 + 298x - 200
\]
\[
P(298) = -\frac{1}{2}(298^2) + 298(298) - 200
\]
\[
P(298) = -\frac{1}{2}(88804) + 88804 - 200
\]
\[
P(298) = -44402 + 88804 - 200
\]
\[
P(298) = 44202
\]

Thus, the maximum profit is:
\[
\boxed{\$44,202}
\]

---

### Final Answers:
(a) Optimal quantity: **298 units**  
(b) Selling price: **\$151 per unit**  
(c) Maximum profit: **\$44,202**

Would you like a detailed breakdown of any step or further clarification?  

---

### Related Questions:
1. How does a change in the demand function affect the profit-maximizing quantity?
2. What happens to the optimal quantity if the cost function changes?
3. How is the break-even quantity calculated for this product?
4. What is the impact of a fixed cost increase on the maximum profit?
5. How does elasticity of demand influence the selling price?

**Tip:** Always double-check your derivative calculations when solving optimization problems!

The weekly demand function for x units of a product sold by only one firm is p = 300 − 1/2x dollars, and the average cost of production and sale is C = 200 + 2x dollars. (a) Find the quantity that will maximize profit. (b) Find the selling price at this optimal quantity. (c) What is the maximum profit?

Discover how to maximize profit for a product given a demand function p = 300 − 1/2x and a cost function C = 200 + 2x. This solution involves optimization techniques, quadratic equations, and profit calculations. Learn the optimal quantity, selling price, and maximum profit with detailed steps. Suitable for high school students in Grades 11-12.

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