OSH/4/Solution
From UBCMATH WIKI

 [3 pts] $E'(t)$ is decreasing for $t<8$ and increasing for $t>8$ with a minimum at $t=8$ (1 pt). At the start of a shift, as suggested in the article referenced in problem statement, the error rate might decrease as a resident learns more about the condition of the patients currently under their care (1 pt). Later during a shift, the error rate might increase due to fatigue (1 pt).
 [1 pt] I think the optimal time for shift change is when $E'(t)$ reaches a minimum at $t=8$.
 [2 pts] The total number of errors at time $t$ is given by
 $E(t) = 4t\frac{1}{2}t^2+\frac{1}{48}t^3$ (1 pt for antiderivative, 1 pt for realizing $E(0)=0$  showing this implicitly is fine).
 [2 pts] The average rate of change is given by
 $A(t) = \frac{E(t)E(0)}{t0} = 4  \frac{1}{2}t + \frac{1}{48}t^2$ (1 pt for correct form, 1 pt for simplifying correctly).
 [3 pts]
Option 1 Find the absolute minimum of $A(t)$.
Finding critical points of $A(t)$, we solve the equation $A'(t)= \frac{1}{2} + \frac{1}{24}t =0$ (1 pt).
Option 2 Find the point along the graph of $E(t)$ at which the secant line connecting the origin and $(t,E(t)$ has the smallest slope.
Ploting the graph of $E(t)$, we can see that this will occur at a value of $t$ for which the tangent line to $E(t)$ goes through the origin (1 pt for making it clear this is what they're doing). The tangent to $E(t)$ at $t=a$ is $y=E'(a)(ta)+E(a)$. This goes through the origin when $0=E'(a)(0a)+E(a)$ or equivalently when ...
 $0=4a+a^2\frac{1}{16}a^3 + 4a\frac{1}{2}a^2+\frac{1}{48}a^3$
 $0=\frac{2}{48}a^3 +\frac{1}{2}a^2$
 $a=0$ or $a=12$ (1 pt). The tangent line at the origin has a slope of 4 (and corresponds to an obviously ridiculous scenario) and the tangent line at 12 has a slope of 1 so the minimum average error rate over one shift occurs when the shift length is chosen to be $t=12$ (1 pt).
 [5 pts] See this desmos page for an interactive plot. Marking: Axes should be labeled $t$ and $E(t)$ with a few values marked for scale (1 pt). The function should be increasing everywhere (1 pt). The inflection point at $(8,32/3)$ should have a zero slope (1 pt) (red dot in Desmos). There should be a secant line (blue line) from the origin to $(12,12)$ (orange dot) whose slope gives the average rate of change over a 12 hour shift (1 pt). The secant line should actually be tangent to the graph at the $(12,12)$ (1 pt).
 [1 pt] My expectation does not match the result, which indicates that a shift should extend 4 hours beyond the lowest error rate. During the period from $t=8$ until $t=12$, even though the resident on the first shift is making more errors than they did previously, they accummulate fewer errors than a new resident just starting their shift at $t=8$ would in their first 4 hours.
Subtotal: 17 pts.

 [6 pts] $S(0)=0$. This means that no cancer patient will survive without the drug treatment.
As $q\to\infty$, we can approximate
 $\dfrac{q}{k_1+q}\approx\dfrac{q}{q}=1$ and $\dfrac{q}{k_2+q}\approx\dfrac{q}{q}=1$ so
Students should: (1 pt) Compute $S(0)$.
 (2 pts) The interpretation of $S(0)$. (1 pt for interpreting $S=0$ and 1 pt for also including an interpretation of $q=0$ in the statement).
 (1 pt) Compute the limit as $q\to\infty$ either by approximating the function for large $q$ or by a more formal limit calculation.
 (2 pts) The interpretation of $\lim_{q\to\infty}S(0)$. (1 pt for interpreting $S\to 0$ and 1 pt for also including an interpretation of $q\to\infty$ in the statement).
 [4 points ]
From (a), we know the general shape of $S(q)$, and that its maximum will occur at a critical point (rather than an endpoint). So, we calculate:
$
S'(q)=\frac{k_1}{(k_1+q)^2}\frac{k_2}{(k_2+q)^2}$
Since $S'(q)$ exists everywhere in our model domain, we find its zeroes.
$\begin{align*}0&=\frac{k_1}{(k_1+q)^2}\frac{k_2}{(k_2+q)^2}\\ \frac{k_1}{(k_1+q)^2}&=\frac{k_2}{(k_2+q)^2}\\ k_1(k_2^2+2k_2q+q^2)&=k_2(k_1^2+2k_1q+q^2)\\ k_1k_2^2+2k_1k_2q+k_1q^2&=k_1^2k_2+2k_1k_2q+k_2q^2\\ k_1k_2(k_2k_1)&=q^2(k_2k_1)\\ k_1k_2&=q^2\\ q&=\sqrt{k_1k_2} \end{align*}$
(In this calculation, we made use of the fact that $k_1 \neq k_2$ and $q \ge 0$.) So, the optimal dosage is $q=\sqrt{k_1k_2}$.
Students should: (1 pt) Compute $S'(q)$.
 (1 pt) Find the critical point of $S(q)$.
 (1 pt) Justify that this is a maximum.
 (1 pt) Interpret this as the optimal dosage.
 [3 pts]
As we found in (b), if $k_1=25$, the survival rate is maximized by a dosage of $q_*(k_2)=\sqrt{25k_2}=5\sqrt{k_2}$. However, the given information is that we must add the constraint $q_*(k_2) \le 45.$ So, the function giving the optimal dose in this case is:
\[q_*(k_2)=\begin{cases}
5\sqrt{k_2} & \text{ if }k_2 \le 81\\
45 & \text{ if }k_2 > 81
\end{cases}\]
So, the graph that gives the optimum dosage, based on the patient's $k_2$ value, is:
2 points: explanation
1 point: graph
Subtotal: 13 pts
 [6 pts] $S(0)=0$. This means that no cancer patient will survive without the drug treatment.
As $q\to\infty$, we can approximate
Communication: 2 pts
Presentation: 1 pt
Total: 33 pts.