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------------------------------------------------ The marketing division of a large firm has found that it can model the sales generated by an advertising campaign as [tex]s(u) = 0.75u + 1.5[/tex] millions of dollars when the firm invests [tex]u[/tex] thousand dollars in advertising. The firm plans to invest [tex]u(x) = -2.3x^2 + 56x + 200[/tex] thousand dollars each month, where [tex]x[/tex] is the number of months after the beginning of the advertising campaign.

How quickly will the sales for this firm be changing when [tex]x = 14[/tex]? (Round your answer to three decimal places.)

To find how quickly sales are changing when x = 14, we need to calculate the derivative of the sales function s(u) with respect to time (ds/dt), which is the product of ds/du and du/dx at x = 14. The composition of these derivatives will provide the rate of sales change at that specific month.

The question asks to determine how quickly sales will be changing for a firm when x = 14 months after the beginning of an advertising campaign. The sales function is given by s(u) = 0.75u + 1.5 million dollars, where u is the investment in thousands of dollars, and u as a function of x (months) is u(x) = -2.3x2 + 56x + 200. To find how quickly sales are changing, we need to find the derivative of the sales function with respect to time (ds/dt) when x = 14, which requires finding the composition of the derivatives ds/du and du/dx.

Firstly, the derivative of u(x) with respect to x is du/dx = -4.6x + 56, and at x = 14, du/dx = -4.6(14) + 56.

Next, since s(u) is linear with respect to u, ds/du = 0.75.

The rate of change of sales with respect to time is the product of these derivatives, so ds/dt = ds/du imes du/dx, and after substituting the values at x = 14, we can round the answer to three decimal places.