OSH/3/Solution
From UBCMATH WIKI
Solution key
Solution
 When $k>4$, $C(x)>M(x)$ for all $x$. This means that for the same amount of time studying, I will do better in chemistry than in math. I would spend more time studying for the harder subject than the easier one.
 My average on the two exams will be $A(x) = (M(x) + C(20x))/2$.
 The mathematical domain is all values of $x$ except for $x=24$ and $x=k$. The model domain is $0\le x \le 20$ because I can't study a negative amount of time and I only have a total of 20 hours in which to study.
 To find the critical points, I find $A'(x)=M'(x)  C'(20x)$ where the minus sign comes from a chain rule. The critical points solve $M'(x)=C'(20x)$. It's easiest to find the derivative of $M(x)$ and $C(x)$ by rewriting them as suggested in the solution to OSH 2:
 $M(x) = 100\left(1  \dfrac{k}{k+x}\right)$
 $M'(x) = 100 \dfrac{k}{(k+x)^2}$
 $C(x) = 100\left(1  \dfrac{4}{4+x}\right)$
 $C'(x) = 100 \dfrac{4}{(4+x)^2}$
 $\dfrac{k}{(k+x)^2} = \dfrac{4}{(24x)^2}$.
 $\dfrac{\sqrt{k}}{k+x} = \pm \dfrac{2}{24x}$
 $\sqrt{k}(24x) = \pm 2(k+x)$
 $24 \sqrt{k}\sqrt{k} x = 2k +2x$ or $24 \sqrt{k}\sqrt{k} x = 2k 2x$
 $x_1=2\sqrt{k}\dfrac{ 12\sqrt{k}}{\sqrt{k}+2 }$ and
 $x_2=2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}2 }$
 We want (a) $x_1 \ge 0$ and (b) $x_1 \le 20$.
(a) happens when $k$ satisfies
 $2\sqrt{k}\dfrac{ 12\sqrt{k}}{\sqrt{k}+2 } \ge 0$.
 $12\sqrt{k} \ge 0$
 $k\le144$.
 $2\sqrt{k}\dfrac{ 12\sqrt{k}}{\sqrt{k}+2 } \le 20$
 $\sqrt{k}(12\sqrt{k}) \le 10(\sqrt{k}+2 )$
 $12\sqrt{k}k \le 10\sqrt{k}+20$
 $k  2\sqrt{k}+20 \ge 0$.
For $x_2$ to be in the domain, we require $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}2 } \ge 0$
 $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}2 } \le 20$
 $(\sqrt{k}+1) + 19 \le 0$

 $M''(x) = 200\dfrac{k}{(k+x)^3}$
 $C''(x) = 800\dfrac{1}{(4+x)^3}$.
 $A''(x)= 100\dfrac{k}{(k+x)^3} 400\dfrac{1}{(24x)^3}$.
 Having it best to spend more time on math means that we want $x_1>10$. This simplifies to
 $k 7\sqrt{k}+10<0$
 $(\sqrt{k}2)(\sqrt{k}5)<0$.
 From $k=0$ up to $k=144$, the absolute maximum is given by the critical point $x_1$. For $k>144$, the absolute maximum is at the $x=0$ endpoint of the model domain because $A(0)=\frac{250}{6}$ and $A(20)=\frac{1000}{k+20}<\frac{1000}{164}=\frac{250}{41}$ for $k>144$.
 If $k<4$ then the math exam is easier to study for than the chem exam and after relatively little time studying for math, your expected mark is close to 100% with additional time making very little difference ($M(x)$ is flat pretty quickly). In this case, it doesn't make sense to focus on math for very long. On the other hand, if $k>25$, math is much harder that chem and the slope of $M(x)$ is low compared to that of chem so spending a lot of time on math doesn't do much good  the marks per hour studied are much higher for chem than for math. This doesn't match my initial expectation because I was hoping to do well on both and giving up on math just because it's really hard didn't even occur to me  I love math!
Desmos demo
Allocation of points
 1 pt  for stating (in some way) that for the same mark more time is required studying math than chemistry, or stating that studying math is relatively more difficult than studying chemistry, with some justification in either case.
1 pt  for stating why they think they should study either more math or more chemistry.  1 pt  for properly determining $A(x)$
 1 pt  for properly determining the mathematical domain.
1 pt  for properly determining the model domain.  5 pts  Getting the correct equation (3 pts  part marks for getting $M'(x)$ (1 pt), and $A'(x)$ (1pt)), calculating $x_1$ (1 pt), calculating $x_2$ (1 pt).
 3 pts  Getting $k<144$ as the first condition (1 pt), stating that $x_1\le 20$ is the second condition and simplifying to a quadratic (1 pt), giving a coherent argument for why this second inequality is always true.
2 pts  Getting $k>4$ as the first condition (1 pt), giving a coherent argument for why this second inequality is always false.  3 pts  Calculating $M''(x)$ (1 pt), $C''(x)$ (1 pt), $A''(x)$ (1 pt).
3 pts  Concluding that $A''(x)<0$ on the model domain (1 pt), concluding that a critical point in the model domain must be a local max (1 pt), and furthermore an absolute max (1 pt).  1 pt  for stating that more time on math means $x_1>10$.
1 pt for correctly determining the range of $k$ for which $x_1>10$.  2 pts  Correct shape from 0 to 144 including zeros at 0 and 144 and rough location of max (1 pt), zero for $k>144$ (1 pt).
 1 pt  for clearly explaining in the case where $k<4$ (don't have to mention $k<4$).
1 pt for clearly explaining in the case where $k>25$ (don't have to mention $k>25$).
Communication: 2 pts
Presentation: 1 pt
Total 30 pts.