Answer
Jason will earn extra $69.72$ dollars if he sells his $70$th car during these $10$ minutes.
Work Step by Step
We have $S\left( {h,c} \right) = 20h{\left( {1 + \frac{c}{{100}}} \right)^{1.5}}$, where $h$ is the number of hours worked and $c$ is the number of cars sold.
The partial derivatives are
${S_h}\left( {h,c} \right) = 20{\left( {1 + \frac{c}{{100}}} \right)^{1.5}}$, ${\ \ \ }$ ${S_c}\left( {h,c} \right) = \frac{3}{{10}}h{\left( {1 + \frac{c}{{100}}} \right)^{0.5}}$
Jason has worked $160$ hours and sold $69$ cars. Write $\left( {a,b} \right) = \left( {160,69} \right)$. If he sells his $70$th car during these $10$ minutes, then the increments are $1$ car and $1/6$ hours
Let $\left( {m,n} \right) = \left( {\frac{1}{6},1} \right)$. So, the extra money Jason will earn can be approximated using the linear approximation:
$\Delta S \approx {S_h}\left( {a,b} \right)m + {S_c}\left( {a,b} \right)n$
$\Delta S \approx {S_h}\left( {160,69} \right)m + {S_c}\left( {160,69} \right)n$
$\Delta S \approx 20{\left( {1 + \frac{{69}}{{100}}} \right)^{1.5}}\cdot\frac{1}{6} + \frac{3}{{10}}\cdot160{\left( {1 + \frac{{69}}{{100}}} \right)^{0.5}}\cdot1$
$\Delta S \approx 69.72$
So, Jason will earn extra $69.72$ dollars if he sells his $70$th car during these $10$ minutes.
