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) \( \mathrm{S}^{(2)}=\square \) \( \mathrm{S}^{\prime}(2)=4.20 \)
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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!