< OSH‎ | 3
Jump to: navigation, search

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. 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.
  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 want (a) $x_1 \ge 0$ and (b) $x_1 \le 20$. (a) happens when $k$ satisfies
    $2\sqrt{k}\dfrac{ 12-\sqrt{k}}{\sqrt{k}+2 } \ge 0$.
    The $2\sqrt{k}$ and the denominator are always positive so this is equivalent to the condition
    $12-\sqrt{k} \ge 0$
    which is true provided
    (b) happens when
    $2\sqrt{k}\dfrac{ 12-\sqrt{k}}{\sqrt{k}+2 } \le 20$
    which is equivalent to
    $\sqrt{k}(12-\sqrt{k}) \le 10(\sqrt{k}+2 )$
    (no inequality flip because $\sqrt{k}+2>0$) which can be rewritten as
    $12\sqrt{k}-k \le 10\sqrt{k}+20$
    $k - 2\sqrt{k}+20 \ge 0$.
    This is a quadratic function of $\sqrt{k}$. Completing the square gives us $(\sqrt{k}-1)^2+19 \ge 0$. This is always a true inequality so $x_1\le 20$ for all $k>0$.
    For $x_2$ to be in the domain, we require
    $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}-2 } \ge 0$
    which happens for $k \ge 4$. However, we also require that
    $2\sqrt{k}\dfrac{ 12+\sqrt{k}}{\sqrt{k}-2 } \le 20$
    Rearranging this inequality as we did for $x_1$ (this time no inequality flip because we want $k>4$ so the denominator is positive when we care about it), we get
    $(\sqrt{k}+1)^2 + 19 \le 0$
    but this is never possible so $x_2$ is never in the domain.
  6. $M''(x) = -200\dfrac{k}{(k+x)^3}$
    $C''(x) = -800\dfrac{1}{(4+x)^3}$.
    Taking the derivative of the original expression for $A'(x)$ above, we get that $A''(x)=(M''(x) + C''(20-x))/2$ and so
    $A''(x)= -100\dfrac{k}{(k+x)^3} -400\dfrac{1}{(24-x)^3}$.
    For any $0\le x \le 20$, we can see that $A''(x)<0$. Any critical point lying in the interval $[0,20]$ must therefore be a local maximum. Because $A'(x)$ is always decreasing in this interval, the function can have at most one such local maximum and the values at the ends of the model domain must also be lower than the maximum. Thus, any local maximum within the model domain is an absolute maximum.
  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. 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 $k>144$.
    The optimal time spent studying math, $x_1$, as a function of the parameter $k$ (roughly speaking, the difficulty of the course).
  9. 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. 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.
  2. 1 pt - for properly determining $A(x)$
  3. 1 pt - for properly determining the mathematical domain.
    1 pt - for properly determining the model domain.
  4. 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).
  5. 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.
  6. 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).
  7. 1 pt - for stating that more time on math means $x_1>10$.
    1 pt -for correctly determining the range of $k$ for which $x_1>10$.
  8. 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).
  9. 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.