I'm not sure about this, can I ask you if I'm wrong? Thanks.

Data: $$Q_{s}=P/2$$ $$Q_{d}=P-42$$


a) How many units will be traded at 35 dollars? At $14 dollars?

b) What quantity at what price will be in equilibrium?

c) What is the total revenue from sales?

Answer a

$$Q_{s}=P/2 => P=2Q_{s}$$ $$Q_{d}=P-42 => P=Q_{d}+42$$ $$P/2=P-42$$ $$P_{eq}=84$$

35 dollars is below 84 dollars, therefore, there is a shortage (we should plug the 35 dollars into the supply equation)

=> 17 units will be sold at that price (am I correct to round down?)

$$Q_{s}=35/2 = 17.5$$

14 dollars also is below 84 dollars, therefore, there is a shortage too (we should plug the 14 dollars into the supply equation)

=> 7 units will be sold at that price

$$Q_{s}=14/2 = 7$$

Answer b

$$P_{eq}=84$$ $$Q_{eq}=P_{eq} /2 = P_{eq}-42$$ $$Q_{eq}= 42$$

Answer c

The total revenue would be, respectively: $$84*42 = 3528$$ $$35*17 = 595$$ $$14*7 = 98$$

I know it's supposed to be simple but not sure if I'm rounding correctly (17.5 to 17) and if those dramatic drops in revenue are to be expected.

Thank you.

  • 4
    $\begingroup$ Check your demand function, it should be decreasing. $\endgroup$ – VARulle Feb 22 at 12:07
  • $\begingroup$ Thanks for your reply. Not sure I follow you. A demand function is always decreasing. Do you mean by that that I'm incorrect somewhere? I would like more assurance, not less so I would be grateful if you can be more specific. 1) I'm not sure if I should use 17 instead of 17.5. Is 17 correct? 2) I only used the supply function to plug my variables as there is a shortage not a surplus. Therefore I never used the demand function here. Am I wrong then? Thanks. $\endgroup$ – Bachir Messaouri Feb 22 at 12:12
  • 4
    $\begingroup$ Your demand is $Q_d=P-42$, which is increasing in price. I guess it should be $42-P$. Then your calculations will change accordingly. And in general you shouldn't round to integers (if not instructed to do so) since units could well be divisible. $\endgroup$ – VARulle Feb 22 at 12:59
  • $\begingroup$ Ok, I got it ! super thanks. Strange because the original data is part of the question. So this would mean there is an issue there. $\endgroup$ – Bachir Messaouri Feb 22 at 13:07

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