To solve this problem, let's proceed step-by-step.

1. Define the profit function

Profit $P$ is defined as:

$P = R(x) - C(x),$

where $R(x)$ is the revenue and $C(x)$ is the cost.

2. Substitute expressions for $R(x)$ and $C(x)$

The revenue $R(x)$ is given as:

$R(x) = 20x - \frac{1}{200}x^2.$

The cost $C(x)$ is given as:

$C(x) = 2x + 8,000.$

Now, substitute into the profit formula:

$P(x) = \left( 20x - \frac{1}{200}x^2 \right) - \left( 2x + 8,000 \right).$

Simplify:

$P(x) = 20x - \frac{1}{200}x^2 - 2x - 8,000.$

Combine like terms:

$P(x) = 18x - \frac{1}{200}x^2 - 8,000.$

3. Express $x$ in terms of $p$

From the demand equation $x = f(p)$, we are given:

$x = 4,000 - 200p.$

4. Express profit as a function of $p$

Substitute $x = 4,000 - 200p$ into $P(x)$:

$P(p) = 18x - \frac{1}{200}x^2 - 8,000.$

First, substitute $x = 4,000 - 200p$ into each term:

1. First term $18x$:

$18x = 18(4,000 - 200p) = 72,000 - 3,600p.$

2. Second term $\frac{1}{200}x^2$:

$\frac{1}{200}x^2 = \frac{1}{200}(4,000 - 200p)^2.$

Simplify $(4,000 - 200p)^2$:

$(4,000 - 200p)^2 = 16,000,000 - 1,600,000p + 40,000p^2.$

Now divide by 200:

$\frac{1}{200}(4,000 - 200p)^2 = \frac{16,000,000 - 1,600,000p + 40,000p^2}{200}.$

Simplify each term:

$\frac{1}{200}(4,000 - 200p)^2 = 80,000 - 8,000p + 200p^2.$

3. Constant term $-8,000$:

This remains unchanged.

Combine all terms:

$P(p) = (72,000 - 3,600p) - (80,000 - 8,000p + 200p^2) - 8,000.$

Simplify:

$P(p) = 72,000 - 3,600p - 80,000 + 8,000p - 200p^2 - 8,000.$

Combine like terms:

$P(p) = -200p^2 + 4,400p - 16,000.$

5. Find the price $p$ that maximizes profit

The profit function is a quadratic equation:

$P(p) = -200p^2 + 4,400p - 16,000.$

For a quadratic $P(p) = ap^2 + bp + c$, the maximum (since $a < 0$) occurs at:

$p = -\frac{b}{2a}.$

Here:

• $a = -200$,
• $b = 4,400$.

Substitute into the formula:

$p = -\frac{4,400}{2(-200)}.$

Simplify:

$p = \frac{4,400}{400} = 11.$

6. Verify and summarize

The price $p$ that maximizes profit is:

$\boxed{p = 11 \text{ dollars}}$

Additional Questions:

1. What is the maximum profit at $p = 11$?
2. How does the revenue $R(x)$ behave as $p$ increases?
3. What happens to the profit if $p$ is close to the endpoints $p = 0$ or $p = 20$?
4. Can we determine the demand $x$ at $p = 11$?
5. How does the cost compare to revenue when $p = 11$?

Tip:

When dealing with quadratic optimization, always check the sign of the leading coefficient to determine if the vertex is a maximum or minimum.