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I was attempting to complete this question myself. However, I can't seem to get the right answer for part b).

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Here's what I did:

Since Profit= Total Revenue - Total Cost ($P=R-C$), I first used the elimination method to find the solution.

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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?

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    $\begingroup$ The last equation should be $\frac{-121}{12}x^2-4700x=0$. Check your work there. And actually you don't want to solve for $x$, which is simply the quantity sold: you are trying to maximize $P$. How could you pick $x$ so that $P$ is largest? $\endgroup$ – Shaun_the_Post May 30 '18 at 0:37
  • $\begingroup$ @Shaun_the_Post I'm sorry, but I don't understand how the last equation is $-\frac{121}{12}x^2$? And I think I know what you mean... $P$ at it's largest must be the vertex, so I should convert to the vertex form? $\endgroup$ – Jenny B May 30 '18 at 0:58
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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): $377125.92$

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  • $\begingroup$ How did you get the answer for profit? I tried subtracting and got $4700x - \frac{119}{12}x^2$. Do I complete the square? $\endgroup$ – Jenny B May 30 '18 at 1:07
  • $\begingroup$ @JennyB If you want to try that way, then you should use $4700x - \frac{121}{12}x^2$ since $(5000x-10x^2)-(300+ \frac1{12}x^2) = 5000x-10x^2-300- \frac1{12}x^2$ and $-10x^2- \frac1{12}x^2 = - \frac{121}{12}x^2$ $\endgroup$ – Henry May 30 '18 at 9:55

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