The total sales of a company (in millions of dollars) \( t \) months from now are given by \( S(t)=0.05 t^{3}+0.4 t^{2}+2 t+9 \). (A) Find \( S^{\prime}(t) \). (B) Find \( S(2) \) and \( S^{\prime}(2) \) (to two decimal places). (C) Interpret \( S(10)=119.00 \) and \( S^{\prime}(10)=25.00 \). (A) \( S^{\prime}(t)=\square \)

To find the derivative \( S^{\prime}(t) \): The function is \( S(t) = 0.05 t^{3} + 0.4 t^{2} + 2t + 9 \). Using the power rule, we differentiate each term: 1. The derivative of \( 0.05 t^{3} \) is \( 0.15 t^{2} \). 2. The derivative of \( 0.4 t^{2} \) is \( 0.8 t \). 3. The derivative of \( 2t \) is \( 2 \). 4. The derivative of \( 9 \) is \( 0 \). Putting it all together: \[ S^{\prime}(t) = 0.15 t^{2} + 0.8 t + 2 \] So, the answer is \( S^{\prime}(t)=0.15 t^{2}+0.8 t+2 \). Now, let's calculate \( S(2) \) and \( S^{\prime}(2) \): 1. **Calculating \( S(2) \)**: \[ S(2) = 0.05(2^{3}) + 0.4(2^{2}) + 2(2) + 9 \] \[ = 0.05(8) + 0.4(4) + 4 + 9 \] \[ = 0.4 + 1.6 + 4 + 9 \] \[ = 15.00 \] 2. **Calculating \( S^{\prime}(2) \)**: \[ S^{\prime}(2) = 0.15(2^{2}) + 0.8(2) + 2 \] \[ = 0.15(4) + 1.6 + 2 \] \[ = 0.6 + 1.6 + 2 \] \[ = 4.20 \] So, \( S(2) = 15.00 \) and \( S^{\prime}(2) = 4.20 \). Next, let’s interpret the values of \( S(10) \) and \( S^{\prime}(10) \): When \( S(10) = 119.00 \), it means that in 10 months, the company's total sales are projected to be $119 million. \( S^{\prime}(10) = 25.00 \) indicates that at that point in time, the sales are increasing at a rate of $25 million per month. This shows strong growth potential for the business!