# 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 $A(x) = (M(x) + C(20-x))/2$.
- I only have a total of 20 hours so the domain is $0\le x \le 20$.
- To find the critical points, I find $A'(x)=M'(x) - C'(20-x)$ where the minus sign comes from a chain rule. The critical points solve $M'(x)=C'(20-x)$. It's easiest to find the derivative of $M(x)$ and $C(x)$ by rewriting them as suggested in the solution to OSH 2:
- $M(x) = 100\left(1 - \dfrac{k}{k+x}\right)$

- $M'(x) = 100 \dfrac{k}{(k+x)^2}$

- $C(x) = 100\left(1 - \dfrac{4}{4+x}\right)$

- $C'(x) = 100 \dfrac{4}{(4+x)^2}$

- $\dfrac{k}{(k+x)^2} = \dfrac{4}{(24-x)^2}$.

- $\dfrac{\sqrt{k}}{k+x} = \pm \dfrac{2}{24-x}$

- $\sqrt{k}(24-x) = \pm 2(k+x)$
- $24 \sqrt{k}-\sqrt{k} x = 2k +2x$ 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 could use the second derivative test: Taking the derivative of the original expression for $A'(x)$ above, we get that $A''(x)=M''(x) + C''(20-x)$. Also,
- $M''(x) = -200\dfrac{k}{(k+x)^3}$
- $C''(x) = -800\dfrac{1}{(4+x)^3}$

- $A''(x)= -200\dfrac{k}{(k+x)^3} -800\dfrac{1}{(24-x)^3}$.

- We want (A) $x_1 \ge 0$ and (B) $x_1 \le 20$.
(A) happens when $k$ satisfies
- $2\sqrt{k}\dfrac{ 12-\sqrt{k}}{\sqrt{k}+2 } > 0$.

- $\sqrt{k}(12-\sqrt{k})>0$

- $12-\sqrt{k}>0$ or $k<144$.

- $2\sqrt{k}\dfrac{ 12-\sqrt{k}}{\sqrt{k}+2 } < 20$

- $\sqrt{k}(12-\sqrt{k})< 10(\sqrt{k}+2 )$

- $12\sqrt{k}-k< 10\sqrt{k}+20$
- $k - 2\sqrt{k}+20>0$.

- $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}-2 } > 0$

- $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}-2 } < 20$

- $(\sqrt{k}+\frac{1}{2}) + 20-\frac{1}{4}<0$

- Having it best to spend more time on math means that we want $x_1>10$. This simplifies to
- $k -7\sqrt{k}+10<0$

- $(\sqrt{k}-2)(\sqrt{k}-5)<0$.

- If $k<<4$ then the math exam is much easier to study for than the chem exam and after very little time, 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 $k>>4$, 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 guess because I was considering the fact that I don't want to fail either exam which is not a consideration in the model.

**Allocation of points**

**1 pt**- for stating 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 domain, given 20 possible study hours**0 pt**- They can calculate the critical points as above or just plug $x_1$ and $x_2$ into $f'(x)$ and show that $f'(x_1)=f'(x_2)=0$. This part is worth 0 pts due to limited marking resources.**0 pt**- This part is worth 0 pts due to limited marking resources.**1 pt**- for determining the values of $k$ for which $x_1$ is in the domain of the model.

**1 pt**- for determining the values of $k$ for which $x_2$ is in the domain of the model.

No work is required if ranges of $k$ are correct, as students may have used a graphing tool.**1 pt**partial mark is given for setting up inequalities $x_1>0$, $x_1<20$, $x_2>0$ and $x_2<20$ with some algebraic mistake.**1 pt**- for stating it is best when $x_1>10$.

**1 pt**-for correctly determining the range of $k$ for which $x_1>10$.**1 pt**- for clearly explaining in the case where $k<<4$.

**1 pt**-for clearly explaining in the case where $k>>4$

**Subtotal 10 pts.**