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To find the maximum profit for the bike company, we first need to determine the profit function \( P(x) \), which is the difference between the revenue \( R(x) \) and the cost \( C(x) \). Given:

\[ R(x) = 325x - 2x^2 \]
\[ C(x) = 4500 + 55x \]

The profit function \( P(x) \) is:
\[ P(x) = R(x) - C(x) = (325x - 2x^2) - (4500 + 55x) \]

Simplifying this, we get:
\[ P(x) = -2x^2 + 325x - 55x - 4500 \]
\[ P(x) = -2x^2 + 270x - 4500 \]

To find the maximum profit, we need to find the vertex of the parabola represented by \( P(x) \). The \( x \)-coordinate of the vertex (and thus the number of bikes sold for maximum profit) can be found using the formula \( x = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the quadratic term and the linear term of \( P(x) \) respectively.

Here, \( a = -2 \) and \( b = 270 \), so:
\[ x = -\frac{270}{2(-2)} = \frac{270}{4} = 67.5 \]

Since the number of bikes sold, \( x \), must be an integer, we will evaluate the profit at \( x = 67 \) and \( x = 68 \) to determine which gives a higher profit.

Calculating \( P(67) \):
\[ P(67) = -2(67)^2 + 270(67) - 4500 \]

Calculating \( P(68) \):
\[ P(68) = -2(68)^2 + 270(68) - 4500 \]

We will now compute these values.

If a bike company has a revenue of R(x) = 325x - 2x^2 and their costs are C(x) = 4500 + 55x for x bikes sold, find the maximum profit. Remember P(x) = R(x) - C(x). Round to the nearest dollar.

Learn how to find the maximum profit for a bike company by using quadratic equations to model revenue and costs. This problem involves algebraic optimization techniques, and key concepts include profit functions, quadratic formulas, and determining the vertex of a parabola. Suitable for high school students.

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