Question

the total time in minutes that the operators at a call center are on the phone with customers t days after the call center goes live is modeled by $f(t) = 5t^3 + 16t + 16$. the total number of customers calling t days after the call center goes live is modeled by $c(t) = 8t^2 + 5t + 5$. the average time spent on the phone per customer is $g(t) = \frac{f(t)}{c(t)}$.
how fast is the average time spent on the phone per customer changing on the tenth day?
(round to two decimal places.)
minutes per customer per day

Explanation:

To find how fast the average time spent on the phone per customer is changing, we need to find the derivative of \( g(t)=\frac{f(t)}{c(t)} \) using the quotient rule. The quotient rule states that if \( g(t)=\frac{u(t)}{v(t)} \), then \( g^\prime(t)=\frac{u^\prime(t)v(t)-u(t)v^\prime(t)}{[v(t)]^2} \).

Step 1: Identify \( u(t) \), \( v(t) \), \( u^\prime(t) \), and \( v^\prime(t) \)

• Let \( u(t) = f(t)=5t^{3}+16t + 16 \), so \( u^\prime(t)=15t^{2}+16 \) (using the power rule: \( \frac{d}{dt}(t^{n})=nt^{n - 1} \))
• Let \( v(t)=c(t)=8t^{2}+5t + 5 \), so \( v^\prime(t)=16t + 5 \) (using the power rule)

Step 2: Apply the quotient rule

\[
g^\prime(t)=\frac{(15t^{2}+16)(8t^{2}+5t + 5)-(5t^{3}+16t + 16)(16t + 5)}{(8t^{2}+5t + 5)^{2}}
\]

Step 3: Evaluate at \( t = 10 \)

First, calculate the numerator:

• Calculate \( (15t^{2}+16)(8t^{2}+5t + 5) \) at \( t = 10 \):
• \( 15(10)^{2}+16=1500 + 16=1516 \)
• \( 8(10)^{2}+5(10)+5=800+50 + 5=855 \)
• \( 1516\times855=1516\times(800 + 50+5)=1516\times800+1516\times50 + 1516\times5=1212800+75800+7580 = 1296180 \)
• Calculate \( (5t^{3}+16t + 16)(16t + 5) \) at \( t = 10 \):
• \( 5(10)^{3}+16(10)+16=5000+160 + 16=5176 \)
• \( 16(10)+5=165 \)
• \( 5176\times165=5176\times(160 + 5)=5176\times160+5176\times5=828160+25880 = 854040 \)
• Subtract the two results: \( 1296180-854040 = 442140 \)

Next, calculate the denominator:

• \( (8t^{2}+5t + 5)^{2} \) at \( t = 10 \):
• \( 8(10)^{2}+5(10)+5=800 + 50+5=855 \)
• \( 855^{2}=731025 \)

Now, \( g^\prime(10)=\frac{442140}{731025}\approx0.605 \) (rounded to three decimal places for intermediate step, then we will round to two decimal places as required)

Wait, let's recalculate the numerator more accurately:

First, expand \( (15t^{2}+16)(8t^{2}+5t + 5) \):
\[

$$\begin{align*} &15t^{2}(8t^{2}+5t + 5)+16(8t^{2}+5t + 5)\\ =&120t^{4}+75t^{3}+75t^{2}+128t^{2}+80t + 80\\ =&120t^{4}+75t^{3}+203t^{2}+80t + 80 \end{align*}$$

\]

Expand \( (5t^{3}+16t + 16)(16t + 5) \):
\[

$$\begin{align*} &5t^{3}(16t + 5)+16t(16t + 5)+16(16t + 5)\\ =&80t^{4}+25t^{3}+256t^{2}+80t + 256t + 80\\ =&80t^{4}+25t^{3}+256t^{2}+336t + 80 \end{align*}$$

\]

Now, subtract the two expanded polynomials:
\[

$$\begin{align*} &(120t^{4}+75t^{3}+203t^{2}+80t + 80)-(80t^{4}+25t^{3}+256t^{2}+336t + 80)\\ =&120t^{4}-80t^{4}+75t^{3}-25t^{3}+203t^{2}-256t^{2}+80t - 336t+80 - 80\\ =&40t^{4}+50t^{3}-53t^{2}-256t \end{align*}$$

\]

Now, evaluate the numerator at \( t = 10 \):
\[

$$\begin{align*} &40(10)^{4}+50(10)^{3}-53(10)^{2}-256(10)\\ =&40\times10000+50\times1000-53\times100 - 2560\\ =&400000+50000 - 5300-2560\\ =&450000-(5300 + 2560)\\ =&450000 - 7860\\ =&442140 \end{align*}$$

\]

Denominator at \( t = 10 \):
\[
(8(10)^{2}+5(10)+5)^{2}=(800 + 50+5)^{2}=855^{2}=731025
\]

So \( g^\prime(10)=\frac{442140}{731025}\approx0.605 \approx0.61 \) (rounded to two decimal places)

Answer:

\( 0.61 \) minutes per customer per day