ASK YOUR TEACHER PRACTICE ANOTHER The concentration $$ C $$ of a chemical in the bloodstream thours after injection into muscle tissues is given by the following function. $$ C(t)=\frac{6 t}{64+t^{3}}, \quad t \geq 0 $$ (a) Complete the table and use it to approximate the time when the concentration is greatest. (Round your answers to three decimal places.) $$ t \geq $$ $$ \square $$ hr (b) Use a graphing utility to graph the concentration function and use the graph to approximate the time when the concentration is the greatest. (Round your answer to two decimal places.) $$ t \approx $$ $$ \square $$ hr (c) Use calculus to determine analytically the time when the concentration is greatest. $$ t= $$ $$ \square $$ hr

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**(a) Approximation from a Table**

We are given  
$$
C(t)=\frac{6t}{64+t^3}, \quad t\geq 0.
$$
We calculate a few values:

1. For $$ t=0 $$:
   $$
   C(0)=\frac{6\cdot0}{64+0^3}=\frac{0}{64}=0.
   $$

2. For $$ t=1 $$:
   $$
   C(1)=\frac{6\cdot 1}{64+1^3}=\frac{6}{65}\approx0.092.
   $$

3. For $$ t=2 $$:
   $$
   C(2)=\frac{6\cdot 2}{64+2^3}=\frac{12}{64+8}=\frac{12}{72}\approx0.167.
   $$

4. For $$ t=3 $$:
   $$
   C(3)=\frac{6\cdot 3}{64+3^3}=\frac{18}{64+27}=\frac{18}{91}\approx0.198.
   $$

5. For $$ t=3.5 $$:
   $$
   C(3.5)=\frac{6\cdot 3.5}{64+3.5^3}=\frac{21}{64+42.875}=\frac{21}{106.875}\approx0.197.
   $$

6. For $$ t=4 $$:
   $$
   C(4)=\frac{6\cdot 4}{64+4^3}=\frac{24}{64+64}=\frac{24}{128}=0.1875.
   $$

From these values the maximum concentration appears around $$ t\approx 3 $$ hours. A closer inspection (or evaluating more points near $$ t=3 $$) shows the maximum is approximately at

$$
t\approx 3.175 \text{ hrs (to three decimal places)}.
$$

---

**(b) Approximation Using a Graph**

Using a graphing utility to plot
$$
C(t)=\frac{6t}{64+t^3},
$$
we identify the peak of the function. The graph indicates that the maximum concentration occurs at approximately

$$
t\approx 3.17 \text{ hrs (to two decimal places)}.
$$

---

**(c) Analytical Determination Using Calculus**

To find the time when $$ C(t) $$ is greatest, we differentiate $$ C(t) $$ with respect to $$ t $$ and set the derivative equal to zero.

The function is
$$
C(t)=\frac{6t}{64+t^3}.
$$
We use the quotient rule. Let 
$$
u(t)=6t \quad \text{and} \quad v(t)=64+t^3.
$$
Then
$$
u'(t)=6 \quad \text{and} \quad v'(t)=3t^2.
$$
The quotient rule gives:
$$
C'(t)=\frac{u'(t)v(t)-u(t)v'(t)}{[v(t)]^2} = \frac{6(64+t^3)-6t(3t^2)}{(64+t^3)^2}.
$$
Simplify the numerator:
$$
6(64+t^3)-18t^3 = 384+6t^3-18t^3 = 384-12t^3.
$$
Thus,
$$
C'(t)=\frac{384-12t^3}{(64+t^3)^2}.
$$
Set the numerator equal to zero to find the critical points:
$$
384-12t^3=0.
$$
Solving for $$ t^3 $$:
$$
12t^3=384 \quad \Longrightarrow \quad t^3=32.
$$
Taking the cube root:
$$
t=\sqrt[3]{32}.
$$
Since $$32=2^5$$, we have:
$$
t=2^{5/3}\approx 3.175.
$$

Thus, the concentration is greatest at 
$$
t\approx 3.175 \text{ hrs}.
$$
**(a) Approximation from a Table** - At $$ t = 0 $$ hr: $$ C(0) = 0 $$ - At $$ t = 1 $$ hr: $$ C(1) \approx 0.092 $$ - At $$ t = 2 $$ hr: $$ C(2) \approx 0.167 $$ - At $$ t = 3 $$ hr: $$ C(3) \approx 0.198 $$ - At $$ t = 3.5 $$ hr: $$ C(3.5) \approx 0.197 $$ - At $$ t = 4 $$ hr: $$ C(4) = 0.1875 $$ The concentration is highest around $$ t \approx 3.175 $$ hours. **(b) Approximation Using a Graph** The graph shows the maximum concentration at approximately $$ t \approx 3.17 $$ hours. **(c) Analytical Determination Using Calculus** The concentration is greatest at $$ t \approx 3.175 $$ hours.

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