To find the production level that maximizes profit given the cost function [tex]\(C(x) = 4050 + 740x + 1.2x^2\)[/tex] and the constant price level [tex]\(p(x) = 2220\)[/tex], we need to follow these steps:
1. Determine the Revenue Function [tex]\(R(x)\)[/tex]:
The revenue function is calculated by multiplying the price per unit by the quantity sold. Since the price is constant at 2220,
[tex]\[
R(x) = p(x) \cdot x = 2220 \cdot x.
\][/tex]
2. Determine the Profit Function [tex]\(P(x)\)[/tex]:
The profit function is the revenue function minus the cost function,
[tex]\[
P(x) = R(x) - C(x) = 2220x - (4050 + 740x + 1.2x^2).
\][/tex]
Simplifying further,
[tex]\[
P(x) = 2220x - 4050 - 740x - 1.2x^2 = (2220x - 740x - 1.2x^2) - 4050 = 1480x - 1.2x^2 - 4050.
\][/tex]
3. Find the Critical Points:
To find where the profit is maximized, we need to take the derivative of [tex]\(P(x)\)[/tex] and set it to zero. The derivative of the profit function is:
[tex]\[
P'(x) = \frac{d}{dx}(1480x - 1.2x^2 - 4050) = 1480 - 2.4x.
\][/tex]
Setting the derivative equal to zero to find the critical points,
[tex]\[
1480 - 2.4x = 0.
\][/tex]
Solving for [tex]\(x\)[/tex],
[tex]\[
2.4x = 1480 \quad \Rightarrow \quad x = \frac{1480}{2.4} = 616.67.
\][/tex]
4. Confirm the Nature of the Critical Point:
To confirm whether this critical point is a maximum, we perform a second derivative test. The second derivative of the profit function is:
[tex]\[
P''(x) = \frac{d}{dx}(1480 - 2.4x) = -2.4.
\][/tex]
Since [tex]\(P''(x) = -2.4\)[/tex] is negative, it indicates that the function [tex]\(P(x)\)[/tex] is concave down at [tex]\(x = 616.67\)[/tex], meaning this point is a maximum.
Therefore, the production level that maximizes profit is
[tex]\[
\boxed{616.67}.
\][/tex]
