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The lotal sales of a company (in millions of dollars) I months from now are given by \( \mathrm{S}(\mathrm{t})=0.051^{3}+0.4 \mathrm{t}^{2}+21+9 \). (A) Find \( \mathrm{S}^{\prime}(t) \). (B) Find \( \mathrm{S}(2) \) and \( \mathrm{S}^{\prime}(2) \) (to two decimal places). (C) Interpret \( \mathrm{S}(10)=119.00 \) and \( \mathrm{S}^{\prime}(10)=25.00 \). (A) \( \mathrm{S}^{\prime}(t)=0.15 t^{2}+0.8 t+2 \) (B) \( \mathrm{S}^{(2)}=\square \) \( \mathrm{S}^{\prime}(2)=4.20 \)

Ask by Turner Bernard. in the United States
Feb 03,2025

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Tutor-Verified Answer

Answer

(A) \( \mathrm{S}^{\prime}(t) = 0.15t^{2} + 0.8t + 2 \) (B) \( \mathrm{S}(2) = 15.00 \) million dollars and \( \mathrm{S}^{\prime}(2) = 4.20 \) million dollars per month (C) At 10 months, total sales are \$119.00 million, and sales are increasing at \$25.00 million per month.

Solution

Let's solve each part of the problem step by step. ### Given: The sales function is: \[ \mathrm{S}(\mathrm{t}) = 0.05 \mathrm{t}^{3} + 0.4 \mathrm{t}^{2} + 2\mathrm{t} + 9 \] where \( \mathrm{S}(\mathrm{t}) \) is in millions of dollars and \( \mathrm{t} \) is the time in months. ### (A) Find \( \mathrm{S}^{\prime}(t) \). To find the first derivative \( \mathrm{S}^{\prime}(t) \), differentiate \( \mathrm{S}(t) \) with respect to \( t \): \[ \mathrm{S}^{\prime}(t) = \frac{d}{dt}\left(0.05 \mathrm{t}^{3} + 0.4 \mathrm{t}^{2} + 2\mathrm{t} + 9\right) = 0.15 \mathrm{t}^{2} + 0.8 \mathrm{t} + 2 \] **Answer:** \[ \mathrm{S}^{\prime}(t) = 0.15 \mathrm{t}^{2} + 0.8 \mathrm{t} + 2 \] ### (B) Find \( \mathrm{S}(2) \) and \( \mathrm{S}^{\prime}(2) \) (to two decimal places). 1. **Calculate \( \mathrm{S}(2) \):** \[ \mathrm{S}(2) = 0.05(2)^3 + 0.4(2)^2 + 2(2) + 9 = 0.05(8) + 0.4(4) + 4 + 9 = 0.4 + 1.6 + 4 + 9 = 15.00 \text{ million dollars} \] 2. **Calculate \( \mathrm{S}^{\prime}(2) \):** \[ \mathrm{S}^{\prime}(2) = 0.15(2)^2 + 0.8(2) + 2 = 0.15(4) + 1.6 + 2 = 0.6 + 1.6 + 2 = 4.20 \text{ million dollars per month} \] **Answer:** \[ \mathrm{S}(2) = 15.00 \text{ million dollars} \] \[ \mathrm{S}^{\prime}(2) = 4.20 \text{ million dollars per month} \] ### (C) Interpret \( \mathrm{S}(10) = 119.00 \) and \( \mathrm{S}^{\prime}(10) = 25.00 \). 1. **Interpretation of \( \mathrm{S}(10) = 119.00 \):** - This means that **10 months from now**, the total sales are projected to be **\$119 million**. 2. **Interpretation of \( \mathrm{S}^{\prime}(10) = 25.00 \):** - This indicates that **at the 10th month**, the sales are **increasing at a rate of \$25 million per month**. **Answer:** - **(A)** \( \mathrm{S}^{\prime}(t) = 0.15\, t^{2} + 0.8\, t + 2 \) - **(B)** \( \mathrm{S}(2) = 15.00 \) million dollars and \( \mathrm{S}^{\prime}(2) = 4.20 \) million dollars per month - **(C)** At 10 months, total sales are projected to be \$119.00 million, and the sales are increasing at a rate of \$25.00 million per month.

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Bonus Knowledge

The correct derivative for \( S(t) \) can be calculated as follows: (A) The function \( S(t) = 0.051t^3 + 0.4t^2 + 21 + 9 \) simplifies to \( S(t) = 0.051t^3 + 0.4t^2 + 30 \). Taking the derivative, we get \( S'(t) = 0.153t^2 + 0.8t \). (B) To find \( S(2) \), plug \( t=2 \) into \( S(t) \): \[ S(2) = 0.051(2^3) + 0.4(2^2) + 30 = 0.051(8) + 0.4(4) + 30 = 0.408 + 1.6 + 30 = 32.008 \] So, \( S(2) \approx 32.01 \) (to two decimal places). For \( S'(2) \): \[ S'(2) = 0.153(2^2) + 0.8(2) = 0.153(4) + 1.6 = 0.612 + 1.6 = 2.212 \] So, \( S'(2) \approx 2.21 \) (to two decimal places). (C) Now, interpreting \( S(10)=119.00 \) means that in 10 months, the company is projected to make 119 million dollars in sales. On the other hand, \( S'(10)=25.00 \) indicates that at the 10-month mark, the sales are increasing at a rate of 25 million dollars per month—great news for the business’s growth!

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