Question

the function $s(t,r) = 40(t - 50)(6 - r)$ gives an ice cream shops daily sales as a function of temperature $t$ (in $^circ$f) and rain $r$ (in inches). find $s_{tr}(90,5)$ (include the appropriate units) and explain what it means.

$s_{tr}(90,5) = -40$ dollars per $^circ$f per inch

explain the meaning of the correct answer from the previous step. select the correct choice below and fill in the answer box within your choice.

\\(\bigcirc\\) a. the temperature decreases at a rate of $\square$ $^circ$f per dollar per inch of rain when the temperature is $90^circ$f and rainfall is $5$.
\\(\bigcirc\\) b. the temperature increases at a rate of $\square$ $^circ$f per dollar per inch of rain when the temperature is $90^circ$f and rainfall is $5$.
\\(\bigcirc\\) c. daily sales decrease at a rate of $\\$\square$ per $1^circ$f change in temperature per inch of rain when the temperature is $90^circ$f and rainfall is $5$.
\\(\bigcirc\\) d. daily sales increase at a rate of $\\$\square$ per $1^circ$f change in temperature per inch of rain when the temperature is $90^circ$f and rainfall is $5$.

Explanation:

Step1: Recall the second partial derivative

To find \( S_{Tr}(90,5) \), we first need to find the first partial derivative with respect to \( T \), then take the partial derivative of that result with respect to \( r \).

First, find \( S_T \), the partial derivative of \( S(T,r) = 40(T - 50)(6 - r) \) with respect to \( T \). Treat \( r \) as a constant. Using the product rule (or just the constant multiple rule since \( (6 - r) \) is constant with respect to \( T \)):

\( S_T=\frac{\partial S}{\partial T}=40(6 - r)\frac{\partial}{\partial T}(T - 50)=40(6 - r)(1)=40(6 - r) \)

Step2: Find the second partial derivative \( S_{Tr} \)

Now, take the partial derivative of \( S_T = 40(6 - r) \) with respect to \( r \). Treat \( T \) as a constant.

\( S_{Tr}=\frac{\partial}{\partial r}(S_T)=\frac{\partial}{\partial r}[40(6 - r)] = 40\frac{\partial}{\partial r}(6 - r)=40(- 1)=- 40 \)

Notice that the second partial derivative \( S_{Tr} \) is a constant (it does not depend on \( T \) or \( r \)), so \( S_{Tr}(90,5)=-40 \).

The units of \( S \) are dollars (since it's daily sales). The units of \( T \) are \( ^\circ\text{F} \) and units of \( r \) are inches.

For the partial derivative \( S_{Tr} \), which is the rate of change of \( S_T \) (which has units of dollars per \( ^\circ\text{F} \)) with respect to \( r \) (units of inches), so the units of \( S_{Tr} \) are \( \frac{\text{dollars per } ^\circ\text{F}}{\text{inch}}=\text{dollars per } ^\circ\text{F per inch} \).

Now, for the meaning: \( S_{Tr} \) represents the rate of change of the rate of change of sales with respect to temperature, with respect to rain. A negative value means that as we increase rain (by 1 inch) and consider the rate of change of sales with respect to temperature, that rate is decreasing. More precisely, when \( T = 90^\circ\text{F} \) and \( r = 5 \) inches, the daily sales decrease at a rate of \( \$40 \) per \( 1^\circ\text{F} \) change in temperature per inch of rain. So the correct option is C, and we fill in 40.

Answer:

For \( S_{Tr}(90,5) \), the value is \(-40\) dollars per \( ^\circ\text{F} \) per inch.

For the multiple - choice question, the correct option is C. Daily sales decrease at a rate of \(\$40\) per \( 1^\circ\text{F}\) change in temperature per inch of rain when the temperature is \( 90^\circ\text{F}\) and rainfall is \( 5\) inches.