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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?

Respuesta :

it is given that p=750000/q^1.5
p'=-1125000q^(-2.5)<0
p and q is inverse of each other as p is decreasing with increasing of q.
and q  take the allowable least value=5000
charge= 750000/(5000)^1.5
=$2.12/lb
maximum revenue= 2.12(5000)= $23320
hope it helps

Answer:

The town't fishery should charges around $2 per pound.

Step-by-step explanation:

Givens:

  • [tex]p=\frac{700,000}{q^{1.5} }[/tex]; where [tex]q[/tex] refers to the number of pounds of tune, and [tex]p[/tex] refers to dollar cost per pound.

So, the problem is asking the total price is the town's fishery sells at leas 5,000 pounds. To solve this problem we just need to replace the given values in the expression:

[tex]p=\frac{700,000}{(5,000)^{1.5} }[/tex]

[tex]p=\frac{700,000}{353,553.4}[/tex]

[tex]p=1.98[/tex]

Therefore, the town't fishery should charges around $2 per pound.