Questions: Quiz 2 (2.3-2.4) - MAC2233 CAI Quiz 2 (2.3-2.4) Question 7 of 10 The total sales of a company (in millions of dollars) t months from now are given by the following formula. S(t)=9 sqrt(t+1) (A) Use the four-step process to find S'(t). (B) Find S(15) and S'(15). (C) Use the results in part (B) to estimate the total sales after 16 months and 17 months. (A) S'(t)= (B) S(15)= (Type an integer or a decimal.) S'(15)= (Type an integer or a decimal.) (C) S(16) approx (Type an integer or a decimal.) S(17) approx (Type an integer or a decimal.)

Quiz 2 (2.3-2.4) - MAC2233 CAI Quiz 2 (2.3-2.4) Question 7 of 10 The total sales of a company (in millions of dollars) t months from now are given by the following formula. S(t)=9 sqrt(t+1) (A) Use the four-step process to find S'(t). (B) Find S(15) and S'(15). (C) Use the results in part (B) to estimate the total sales after 16 months and 17 months. (A) S'(t)= (B) S(15)= (Type an integer or a decimal.) S'(15)= (Type an integer or a decimal.) (C) S(16) approx (Type an integer or a decimal.) S(17) approx (Type an integer or a decimal.)
Transcript text: Quiz 2 (2.3-2.4) - MAC2233 CAI Quiz 2 (2.3-2.4) Question 7 of 10 The total sales of a company (in millions of dollars) $t$ months from now are given by the following formula. \[ S(t)=9 \sqrt{t+1} \] (A) Use the four-step process to find $\mathrm{S}^{\prime}(\mathrm{t})$. (B) Find $\mathrm{S}(15)$ and $\mathrm{S}^{\prime}(15)$. (C) Use the results in part (B) to estimate the total sales after 16 months and 17 months. (A) $S^{\prime}(t)=$ $\square$ (B) $\mathrm{S}(15)=$ $\square$ (Type an integer or a decimal.) $S^{\prime}(15)=$ $\square$ (Type an integer or a decimal.) (C) $\mathrm{S}(16) \approx$ $\square$ (Type an integer or a decimal.) $\mathrm{S}(17) \approx$ $\square$ (Type an integer or a decimal.)
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Solution

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Solution Steps

Step 1: Find \( S'(t) \)

To find the derivative \( S'(t) \) of the function \( S(t) = 9 \sqrt{t+1} \), we apply the chain rule. The derivative is given by: \[ S'(t) = 9 \cdot \frac{1}{2\sqrt{t+1}} = \frac{9}{2\sqrt{t+1}} \]

Step 2: Calculate \( S(15) \) and \( S'(15) \)

Substituting \( t = 15 \) into the function \( S(t) \): \[ S(15) = 9 \sqrt{15 + 1} = 9 \sqrt{16} = 36.0 \] Now, substituting \( t = 15 \) into the derivative: \[ S'(15) = \frac{9}{2\sqrt{15 + 1}} = \frac{9}{2\sqrt{16}} = \frac{9}{8} = 1.125 \]

Step 3: Estimate \( S(16) \) and \( S(17) \)

Using the linear approximation: \[ S(16) \approx S(15) + S'(15) \cdot 1 = 36.0 + 1.125 = 37.125 \] \[ S(17) \approx S(16) + S'(15) \cdot 1 = 37.125 + 1.125 = 38.25 \]

Final Answer

\[ S(15) = 36.0, \quad S'(15) = 1.125, \quad S(16) \approx 37.125, \quad S(17) \approx 38.25 \] Thus, the final answers are: \[ \boxed{S(15) = 36.0} \] \[ \boxed{S'(15) = 1.125} \] \[ \boxed{S(16) \approx 37.125} \] \[ \boxed{S(17) \approx 38.25} \]

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