Let's break down this profit problem into its parts.

Part (a): Express the profit function $P(x)$

The profit function $P(x)$ is the difference between the revenue $R(x)$ and the cost $C(x)$:

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

Revenue Function $R(x)$:

The revenue $R(x)$ is given by the price $p$ multiplied by the number of units sold $x$. From the problem, we know the price function:

$p = -0.05x + 38$

Since $x$ represents hundreds of units, the total revenue is:

$R(x) = p \cdot x = (-0.05x + 38) \cdot x$

Simplifying:

$R(x) = -0.05x^2 + 38x$

Cost Function $C(x)$:

The cost function is already given as:

$C(x) = 0.02x^2 + 3x + 574.77$

Profit Function $P(x)$:

Now, the profit is:

$P(x) = R(x) - C(x) = (-0.05x^2 + 38x) - (0.02x^2 + 3x + 574.77)$

Simplifying:

$P(x) = -0.05x^2 + 38x - 0.02x^2 - 3x - 574.77$

$P(x) = -0.07x^2 + 35x - 574.77$

So the profit function is:

$P(x) = -0.07x^2 + 35x - 574.77$

Sketching the Graph:

This is a quadratic function (a parabola) that opens downward because the coefficient of $x^2$ is negative. The vertex of this parabola represents the maximum profit.

Part (b): Find the average profit when the price is $37 per unit

First, we need to determine the level of production $x$ when the price is $37. Using the price function $p = -0.05x + 38$:

$37 = -0.05x + 38$

Solving for $x$:

$37 - 38 = -0.05x$ $-1 = -0.05x$ $x = 20$

So, when the price is $37 per unit, $x = 20$ hundred units are produced, or 2000 units.

The average profit $AP$ is the total profit divided by the number of units:

$AP = \frac{P(x)}{x}$

Using $x = 20$ in the profit function:

$P(20) = -0.07(20)^2 + 35(20) - 574.77$ $P(20) = -0.07(400) + 700 - 574.77$ $P(20) = -28 + 700 - 574.77$ $P(20) = 97.23$

Thus, the total profit when $x = 20$ is 97.23 hundred dollars, or $9,723. The average profit is:

$AP = \frac{P(20)}{20} = \frac{97.23}{20} = 4.8615 \text{ hundred dollars, or } \$486.15$

Part (c): Maximum Profit

To find the level of production $x$ that maximizes profit, we take the derivative of $P(x)$ and set it equal to zero:

$P'(x) = -0.14x + 35$ $0 = -0.14x + 35$ $0.14x = 35$ $x = \frac{35}{0.14} = 250$

So, the level of production that maximizes profit is $x = 250$ hundred units, or 25,000 units.

Price corresponding to maximum profit:

Using the price function $p = -0.05x + 38$ to find the price at $x = 250$:

$p = -0.05(250) + 38 = -12.5 + 38 = 25.5$

So, the unit price that corresponds to maximum profit is $25.50.

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Summary:

1. Profit function: $P(x) = -0.07x^2 + 35x - 574.77$
2. Average profit at price $37: $486.15
3. Maximum profit occurs when $x = 250$ hundred units (25,000 units) are produced, with a unit price of $25.50.

Would you like me to sketch the graph of the profit function or clarify any details?

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Related questions:

1. What happens to the profit if the unit price increases beyond $38?
2. How does the cost function impact the maximum profit?
3. What is the breakeven point for production in this scenario?
4. How sensitive is the profit to small changes in production levels around the maximum?
5. What would happen to the profit function if fixed costs increased?

Tip:

For maximizing profit, always look for the vertex of the profit function if it is quadratic.