Question

Assume that the demand for tuna in a small coastal town is given by

p = 700,000/ q^1.5 ; where q is the number of pounds of tuna that can be sold in a month at p dollars per pound. Assume that the town's fishery wishes to sell at least 5,000 pounds of tuna per month.

(a) How much should the town's fishery charge for tuna in order to maximize monthly revenue?

Answer 1

The town is looking to maximize its revenue from tuna, given by the quantity sold times the price. Hence, revenue r = p*q. But we have a function for p in terms of q, so let's rewrite this:\[r = q*(700000/q^{1.5}) = 700000/q^{0.5} = 700000*q^{-0.5}\]Functions take on maximum values at either places where the value of their derivative is 0 or at the endpoints of the range under consideration. Let's take the derivative of our revenue function:\[r' = -0.5*700000*q^{-1.5} = -350000*q^{-1.5}\]We want to find the zeroes of this function, so set it equal to zero and solve:\[0 = -350000*q^{-1.5}\]Clearly this is only going to be 0 at q = 0. Then we need to look at this point and the endpoints of our range (minimum of 5000, maximum of positive infinity).
r(0) = 700000*0 = 0
r(5000) = 700000*0.014121 = 9899.47
r(infinity) = 700000*0 = 0
Thus, the maximum revenue the town can receive occurs at the sale of 5000 lb of tuna.


As a side note, we can see from the revenue function that this is an inevitable result. Our revenue function was inversely proportional to q; the larger q got, the less money the town made, so the smallest amount of tuna we can get away with selling will be the best.

Answer 2

@Blacksteel wow thanks a lot for your help. can you tell me what step i should do if i wanted to find out this questions?

Answer 3

b) How much tuna will it sell per month at that price?

Answer 4

I can, but I'd like you to check something first - I just realized that's a REALLY weird demand function. Is it supposed to be p pounds of tuna at q dollars per month?

Answer 5

(I'd feel really bad if I gave you the wrong answer because of that, and it seems unlikely that we'd be using a demand function where you could sell one pound of tuna for 700,000 dollars. Either that or I have the wrong job.)

Answer 6

where q is the number of pounds of tuna that can be sold in a month at p dollars per pound

Answer 7

i just read your work and i submitted your answer which is 500, unfortunely it was incorrect. do you know what the problem is?

Answer 8

*shrug* Okay then. I'm not sure why the town wouldn't just sell one pound of tuna for $700k then, but I'm not an expert fisherman.

In that case, we actually already solved question B and kind of forgot to solve question A. We showed that the town will do best selling 5000 lb of tuna, so it will sell at the price:
p = 700000/(5000^1.5) = $1.9799/lb

For part B, it should be 5000, which we found above.

Answer 9

And I very strongly suspect the problem is a typo in the question, since that demand function makes no sense and a function with price q and poundage p would be a normal demand function.

Answer 10

they wanted me to round the answer to the nearest cent. maybe that helps?

Answer 11

Well, if the numbers are right,
A.) 1.98
B.) 5000

Answer 12

how do i find the total revenue?

Answer 13

Total revenue will be pounds sold times price per pound, so r = p*q = 5000 * 1.98 = 9900.

Answer 14

@Blacksteel thanks a lot for the help. i really appreciate it

Answer 15

No problem, happy to help.