I was attempting to complete this question myself. However, I can't seem to get the right answer for part b).
Here's what I did:
Since Profit= Total Revenue - Total Cost ($P=R-C$), I first used the elimination method to find the solution.
The problem here is, I'm stuck on isolating for $x$. I tried the quadratic formula:
$x= -(-4700) \pm \frac{\sqrt (-4700)^2-4(\frac{121}{12})(0)}{\frac {121}{12}}$
= $4700 \pm \frac{\sqrt 22090000}{\frac{121}{6}}$
I didn't get the correct answer, which is $377 125.92.
I also considered factorization, but I'm pretty sure you can't factor this.
(I tried googling it and found this forum: http://mymathforum.com/algebra/39004-systems-solving-inequalities-two-variables.html
However, I'm confused as to how can you complete the square?)
What should I be doing?
The question looks wrong to me: $300+ \frac14 x^2$ is the derivative of $300x+ \frac1{12} x^3$ rather than of $300x+ \frac1{12} x^2$. So any conclusions should be taken with a pinch of salt.
By taking liberties, it is possible to get the answer you quote.
If you set $C_M=R_M$ then you get $$4700 - 20x - \frac14 x^2=0$$ which has the non-negative solution $x=20(\sqrt{51}-2) \approx 102.8285686$
Apparently you then need to round this to the optimal integer $103$ and saying total profit is revenues minus costs find the corresponding value of the erroneous $R-C=(5000x-10x^2)-(300+ \frac1{12}x^2)$ which would be (rounded to two decimal places): $377125.92$