Questions: An artist is planning to sell signed prints of her latest work. If 30 prints are offered for sale, she can charge 450 each. However, if she makes more than 30 prints, she must lower the price of all the prints by 5 for each print in excess of the 30. That is, 31 prints reduce the price by 5, 32 prints reduce the price by 10, and so on. How many prints should the artist make to maximize her revenue?

Solution Steps

To maximize the artist's revenue, we need to express the revenue as a function of the number of prints and then find the number of prints that maximizes this function. The revenue is calculated as the number of prints multiplied by the price per print. The price per print decreases by $5 for each print over 30. We can set up a quadratic equation for revenue and find its maximum by determining the vertex of the parabola.

The revenue \( R \) generated by selling \( n \) prints can be expressed as: \[ R(n) = \begin{cases} 450n & \text{if } n \leq 30 \\ n(450 - 5(n - 30)) & \text{if } n > 30 \end{cases} \] For \( n > 30 \), this simplifies to: \[ R(n) = n(450 - 5n + 150) = n(600 - 5n) = 600n - 5n^2 \]

The revenue function \( R(n) = 600n - 5n^2 \) is a quadratic function that opens downwards (since the coefficient of \( n^2 \) is negative). The maximum revenue occurs at the vertex of the parabola, which can be found using the formula: \[ n = -\frac{b}{2a} \] where \( a = -5 \) and \( b = 600 \). Thus, \[ n = -\frac{600}{2 \times -5} = \frac{600}{10} = 60 \]

Substituting \( n = 60 \) back into the revenue function: \[ R(60) = 600(60) - 5(60^2) = 36000 - 18000 = 18000 \]

Final Answer