Maximizing Ruby Revenue: Price for Optimal Sales

[tony873004]

The demand for rubies is given by the equation
$$q = \frac{4}{3}p + 80$$

where p is the price and q is the number of rubies sold each week. At what price should the rubies be sold to maximize weekly revenue?

$$\begin{array}{l} R = pq \\ R = p\left( {\frac{4}{3}p + 80} \right) \\ R = \frac{4}{3}p^2 + 80p \\ \\ R' = \frac{8}{3}p + 80 \\ \\ \frac{8}{3}p + 80 = 0 \\ \\ \frac{8}{3}p = - 80 \\ \\ p = \frac{{ - 80}}{{\left( {\frac{8}{3}} \right)}} = \frac{{ - 240}}{8} = - 30 \\ \end{array}$$
To maximize weekly revenue, they should give away the rubies and $30 per rubie. (Obviously wrong. The back of the book says $30)

 

[vsage]

It looks like by the second derivative of R with respect to P that actually minimizes the revenue. To maximize revenue then would be to choose the highest price possible. The demand formula doesn't make sense to begin with because it implies that you sell more rubies as the price goes up on rubies, so a strange answer doesn't seem out of the question.

 

[Lyuokdea]

This doesn't make any sense.

q = 4/3p + 80

So when you increase the price you sell more rubies? In that case you would maximize weekly revenue by selling an infinite number of rubies. Either I'm not understanding you correctly, or you are missing a minus sign in that formula, try something like:

q = -4/3p + 80

and see how that works.

EDIT: That seems to give you the right answer, it must be a typo in the book.

~Lyuokdea

 

[dav2008]

To go along with what Lyokdea said, you also know something is wrong when you look at your equation for revenue that you're trying to maximize: R=4/3p²+80p. This function clearly has no absolute maximum; only a minimum. (It's an upwards-opening parabola)

 

[nrqed]

The demand for rubies is given by the equation
$$q = \frac{4}{3}p + 80$$

where p is the price and q is the number of rubies sold each week. At what price should the rubies be sold to maximize weekly revenue?

$$\begin{array}{l} R = pq \\ R = p\left( {\frac{4}{3}p + 80} \right) \\ R = \frac{4}{3}p^2 + 80p \\ \\ R' = \frac{8}{3}p + 80 \\ \\ \frac{8}{3}p + 80 = 0 \\ \\ \frac{8}{3}p = - 80 \\ \\ p = \frac{{ - 80}}{{\left( {\frac{8}{3}} \right)}} = \frac{{ - 240}}{8} = - 30 \\ \end{array}$$
To maximize weekly revenue, they should give away the rubies and $30 per rubie. (Obviously wrong. The back of the book says $30)

:rofl: :rofl: You made me laugh out loud with your last sentence

Are you sure that the first equation should not read -4/3p+80?
With the equation you gave, increasing the price keeps increasing the number sold, which is weird. And notice that your function for R is concave upward..Taking the price to infinity will lead to infinite revenues!
(what you found is actually the minimum).

If the equation for R was -4/3p + 80 it would fix everything and make sense now. Maybe a typo in the book?

Patrick

 

[tony873004]

sorry, I was blind and didn't see the negative sign. -4/3p + 80. Sorry for the inconvenience.

 

[Lyuokdea]

If the euqation for R was -3/4p + 80 it would fix everything and make sense now. Maybe a typo in the book?

nrged, you mean -4/3p + 80, not -3/4p right? That's what came out to match his answer when I ran through the math.

~Lyuokdea

 

[tony873004]

To go along with what Lyokdea said, you also know something is wrong when you look at your equation for revenue that you're trying to maximize: R=4/3p²+80p. This function clearly has no absolute maximum; only a minimum. (It's an upwards-opening parabola)

I really need to learn to spot stuff like this rather than just diving straight into the formulas.

 

[nrqed]

nrged, you mean -4/3p + 80, not -3/4p right? That's what came out to match his answer when I ran through the math.

~Lyuokdea

Yes, absolutely. I will correct it. That was a typo. Thanks for letting me know.
And sorry for having repeated what you said, I just saw your post after I sent mine.

Regards

 

[Curious3141]

If you were a jeweller giving away free rubies plus $30, I'll take whatever you got. :rofl: