2. ( 10 marks) Turkey embryos The oxygen consumption of a turkey embryo changes from the time the egg is laid until the time the turkey chick hatches. In a brush turkey the oxygen consumption (in millilitres per hour) can be modeled by \[ c(t)=-0.0012 t^{3}+0.12 t^{2}-1.83 t+3.97 \] where \( t \) is the time in days since the egg was laid (turkey chicks usually hatch around day 50 ). a) Is the oxygen consumption increasing or decreasing on day 5 ? At what rate? b) Is the oxygen consumption increasing or decreasing on day 20 ? At what rate? c) When between days 0 and 50 is the oxygen consumption increasing at the fastest rate? What is that rate of increase?

**Step 1. Find the rate of change of oxygen consumption, \( c'(t) \).** The given function is \[ c(t)=-0.0012t^3+0.12t^2-1.83t+3.97. \] Differentiate term‐by‐term with respect to \( t \): \[ \begin{aligned} c'(t) &= \frac{d}{dt}\Bigl[-0.0012t^3\Bigr] + \frac{d}{dt}\Bigl[0.12t^2\Bigr] + \frac{d}{dt}\Bigl[-1.83t\Bigr] + \frac{d}{dt}\Bigl[3.97\Bigr] \\ &= -0.0036t^2 + 0.24t - 1.83. \end{aligned} \] --- **Step 2. Answer part (a): Behavior on day 5.** Substitute \( t=5 \) into \( c'(t) \): \[ c'(5) = -0.0036(5)^2 + 0.24(5) - 1.83. \] Calculate each term: - Compute \( (5)^2 = 25 \). - Then, \(-0.0036 \times 25 = -0.09\). - Next, \(0.24 \times 5 = 1.20\). Thus, \[ c'(5) = -0.09 + 1.20 - 1.83 = -0.72. \] Since \( c'(5) \) is negative, the oxygen consumption is **decreasing** on day 5 at a rate of \(-0.72\) millilitres per hour. --- **Step 3. Answer part (b): Behavior on day 20.** Substitute \( t=20 \): \[ c'(20) = -0.0036(20)^2 + 0.24(20) - 1.83. \] Calculate each term: - Compute \( (20)^2 = 400 \). - Then, \(-0.0036 \times 400 = -1.44\). - Next, \(0.24 \times 20 = 4.80\). Thus, \[ c'(20) = -1.44 + 4.80 - 1.83 = 1.53. \] Since \( c'(20) \) is positive, the oxygen consumption is **increasing** on day 20 at a rate of \(1.53\) millilitres per hour. --- **Step 4. Answer part (c): When is the oxygen consumption increasing at the fastest rate between days 0 and 50?** The oxygen consumption is increasing at the fastest rate when \( c'(t) \) is maximized. To find the maximum of \( c'(t) = -0.0036t^2 + 0.24t - 1.83 \), we find its critical point by setting the derivative of \( c'(t) \) equal to 0. Compute the second derivative, \( c''(t) \): \[ c''(t) = \frac{d}{dt}\Bigl[-0.0036t^2+0.24t-1.83\Bigr] = -0.0072t + 0.24. \] Set \( c''(t) = 0 \) to find the critical point: \[ -0.0072t + 0.24 = 0 \quad \Longrightarrow \quad 0.0072t = 0.24 \quad \Longrightarrow \quad t = \frac{0.24}{0.0072}. \] Calculate \( t \): \[ t = 33.\overline{3} \quad \text{days}. \] Since \( c'(t) \) is a quadratic function with a negative coefficient on \( t^2 \), it opens downward. Hence, the critical point at \( t \approx 33.33 \) days is a maximum. Now, compute the maximum rate of increase: \[ c'\left(33.\overline{3}\right) = -0.0036\left(33.\overline{3}\right)^2 + 0.24\left(33.\overline{3}\right) - 1.83. \] To use exact values, express \( t \) as \( \frac{100}{3} \): \[ \begin{aligned} c'\left(\frac{100}{3}\right) &= -0.0036\left(\frac{100}{3}\right)^2 + 0.24\left(\frac{100}{3}\right) - 1.83 \\ &= -0.0036\left(\frac{10000}{9}\right) + \frac{24}{3} - 1.83 \\ &= -\frac{36}{9} + 8 - 1.83 \\ &= -4 + 8 - 1.83 \\ &= 2.17 \quad (\text{approximately}). \end{aligned} \] Thus, between days 0 and 50, the oxygen consumption is increasing at the fastest rate on approximately day \(33.33\), with a rate of increase of about \(2.17\) millilitres per hour.