Step 1 :Given the revenue function, \(R(x) = 7x - 3x^{2}\), and the cost function, \(C(x) = x^{3} - 5x^{2} + 2x + 1\), we need to find the maximum profit and the number of units that must be produced and sold to yield this maximum profit.
Step 2 :The profit function, \(P(x)\), is given by the difference between the revenue function, \(R(x)\), and the cost function, \(C(x)\). That is, \(P(x) = R(x) - C(x)\).
Step 3 :To find the maximum profit, we need to find the maximum of the profit function. This can be done by finding the derivative of the profit function, setting it equal to zero, and solving for x. The values of x that satisfy this equation are the critical points of the profit function.
Step 4 :We then evaluate the profit function at these critical points and at the endpoints of the domain (if any) to find the maximum profit. The number of units that must be produced and sold to yield the maximum profit is the value of x that gives the maximum profit.
Step 5 :The derivative of the profit function is \(-3x^{2} + 4x + 5\). Setting this equal to zero and solving for x gives the critical points \(\frac{2}{3} - \frac{\sqrt{19}}{3}\) and \(\frac{2}{3} + \frac{\sqrt{19}}{3}\).
Step 6 :Evaluating the profit function at these critical points gives the profits \(-\frac{5\sqrt{19}}{3} - \left(\frac{2}{3} - \frac{\sqrt{19}}{3}\right)^{3} + 2\left(\frac{2}{3} - \frac{\sqrt{19}}{3}\right)^{2} + \frac{7}{3}\) and \(-\left(\frac{2}{3} + \frac{\sqrt{19}}{3}\right)^{3} + \frac{7}{3} + \frac{5\sqrt{19}}{3} + 2\left(\frac{2}{3} + \frac{\sqrt{19}}{3}\right)^{2}\).
Step 7 :The maximum profit is \(\boxed{-\left(\frac{2}{3} + \frac{\sqrt{19}}{3}\right)^{3} + \frac{7}{3} + \frac{5\sqrt{19}}{3} + 2\left(\frac{2}{3} + \frac{\sqrt{19}}{3}\right)^{2}}\) thousand dollars and the number of units that must be produced and sold in order to yield the maximum profit is \(\boxed{\frac{2}{3} + \frac{\sqrt{19}}{3}}\) thousand units.