Question

Assume that the demand for tuna in a small coastal town is given by p = 400,000 q1.5 , 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 to maximize monthly revenue? HINT [See Example 3, and don't neglect endpoints.] (Round your answer to the nearest cent.) p = $ per lb (b) How much tuna will it sell per month at that price? q = lb (c) What will be its resulting revenue? (Round your answer to the nearest dollar.) $ per month

Answer 1

(a) p = $ per lb

p = $2.12 per lb

(b) q = lb (c)

q = $23320

Step-by-step explanation:

p=750000/q^1.5=>

p'=-1125000q^(-2.5)<0 always

=>p is decreasing with the

increasing of q. So q should take

the allowable least value=5000.

=>

(a) the charge= 750000/(5000)^1.5= $2.12/lb

(b)The max. revenue=

q = lb (c) = 2.12(5000)= $23320