< OSH‎ | 3
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Solution key


  1. 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.
  2. My average on the two exams will be $A(x) = (M(x) + C(20-x))/2$.
  3. I only have a total of 20 hours so the domain is $0\le x \le 20$.
  4. 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:
    $M(x) = 100\left(1 - \dfrac{k}{k+x}\right)$
    $M'(x) = 100 \dfrac{k}{(k+x)^2}$
    $C(x) = 100\left(1 - \dfrac{4}{4+x}\right)$
    $C'(x) = 100 \dfrac{4}{(4+x)^2}$
    As stated above but with these derivatives plugged in, the critical points solve
    $\dfrac{k}{(k+x)^2} = \dfrac{4}{(24-x)^2}$.
    Taking square roots (recall that $a^2=b^2$ means that $a=\pm b$),
    $\dfrac{\sqrt{k}}{k+x} = \pm \dfrac{2}{24-x}$
    $\sqrt{k}(24-x) = \pm 2(k+x)$
    $24 \sqrt{k}-\sqrt{k} x = 2k +2x$      or      $24 \sqrt{k}-\sqrt{k} x = -2k -2x$
    Thus, the critical points are
    $x_1=2\sqrt{k}\dfrac{ 12-\sqrt{k}}{\sqrt{k}+2 }$ and
    $x_2=2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}-2 }$
  5. 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,
    $M''(x) = -200\dfrac{k}{(k+x)^3}$
    $C''(x) = -800\dfrac{1}{(4+x)^3}$
    $A''(x)= -200\dfrac{k}{(k+x)^3} -800\dfrac{1}{(24-x)^3}$.
    Because in this part we are considering only critical points satisfying $0<x<20$, the denominators in both terms are positive which ensures that $A''(x)<0$ for all $0<x<20$. This means any such critical point is a local maximum. Furthermore, if this local maximum were not the absolute maximum, there would have to be a higher point on the function somewhere in the interval $[0,20]$. But $A(x)$, $A'(x)$ and $A''(x)$ are all continuous on that interval so to get from our local maximum to a point higher up, the concavity would have to switch to positive first which it can't do because we just showed $A''(x)<0$ on $(0,20)$. So there can only be one local maximum and that must be the absolute maximum on $[0,20]$. Alternately, we could also sketch the original function. This function is a sum of a Michaelis-Menten- type saturating function and a second copy of such a function that is reflected about the line $x=20$. The sum of such functions produces a result that has one local maximum.
  6. We want (A) $x_1 \ge 0$ and (B) $x_1 \le 20$. (A) happens when $k$ satisfies
    $2\sqrt{k}\dfrac{ 12-\sqrt{k}}{\sqrt{k}+2 } > 0$.
    The 2 and the denominator are always positive so this is equivalent to the condition
    which is true provided
    $12-\sqrt{k}>0$ or $k<144$.
    (B) happens when
    $2\sqrt{k}\dfrac{ 12-\sqrt{k}}{\sqrt{k}+2 } < 20$
    which is equivalent to
    $\sqrt{k}(12-\sqrt{k})< 10(\sqrt{k}+2 )$
    which can be rewritten as
    $12\sqrt{k}-k< 10\sqrt{k}+20$
    $k - 2\sqrt{k}+20>0$.
    This is a quadratic function of $\sqrt{k}$. Completing the square gives us $(\sqrt{k}-1)^2+19>0$. This is always a true inequality so $x_1<20$ for all $k>0$. For $x_2$ to be in the domain, we require
    $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}-2 } > 0$
    which happens for $k>4$. However, we also require that
    $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}-2 } < 20$
    Rearranging this inequality as we did for $x_1$, we get
    $(\sqrt{k}+\frac{1}{2}) + 20-\frac{1}{4}<0$
    but this is never possible so $x_2$ is never in the domain.
  7. Having it best to spend more time on math means that we want $x_1>10$. This simplifies to
    $k -7\sqrt{k}+10<0$
    which factors into
    For a concave-up parabola, the function is negative between the zeros so it is best to spend more time studying math when $4<k<25$.
  8. 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. 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.
  2. 1 pt - for properly determining $A(x)$
  3. 1 pt - for properly determining the domain, given 20 possible study hours
  4. 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.
  5. 0 pt - This part is worth 0 pts due to limited marking resources.
  6. 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.
  7. 1 pt - for stating it is best when $x_1>10$.
    1 pt -for correctly determining the range of $k$ for which $x_1>10$.
  8. 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.