A small company's profits are related to the price of its single product. The relationship is R(p)=-0.4p^2+64p-2400, where R is the income in thousands of dollars and p is the price of the product in dollars. What price would maximize revenue?

To find the price that maximizes revenue, we need to find the maximum point of the given quadratic function R(p) = -0.4p^2 + 64p - 2400. We can do this by finding the critical points of the function, which occur where the derivative is equal to zero or undefined. To find the derivative, we use the power rule: R'(p) = -0.8p + 64 Setting the derivative equal to zero and solving for p, we get: -0.8p + 64 = 0 -0.8p = -64 p = 80 Now, we need to check if this critical point is a maximum or a minimum by using the second derivative test. The second derivative is: R''(p) = -0.8 Since R''(p) is negative for all values of p, the critical point p = 80 is a maximum point. Therefore, the price that maximizes revenue is $80. To confirm, we can substitute p = 80 into the original function: R(80) = -0.4(80)^2 + 64(80) - 2400 R(80) = -2560 + 5120 - 2400 R(80) = 160 So, the maximum revenue is $160,000 when the price is $80.