In a fishery the long-run harvest function (harvest volume) is H(E) = aE – bE2 a, b positive constants, E is fishing effort. Total cost is TC (E)= cE, with c=unit cost of effort, Total revenue is TR(E)=p*H(E) with p = constant price of fish. a) Find the open-access equilibrium values of effort and harvest, E, and H. , respectively pH(E) ра — с AR(E) = = pa – pbE = MC = c → E. = E pb b) Find the fishing effort that maximizes resource rent, EMEY, and the corresponding harvest, HMEY. What happens to EMEY and HMEY if p increases?

ENGR.ECONOMIC ANALYSIS
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Chapter1: Making Economics Decisions
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In a fishery the long-run harvest function (harvest volume) is
H(E) = aE – bE²
a, b positive constants, E is fishing effort. Total cost is TC (E)= cE, with c=unit cost of effort, Total revenue is TR(E)=p*H(E),
with p = constant price of fish.
a) Find the open-access equilibrium values of effort and harvest, E, and Ho , respectively
pH(E)
ра — с
AR(E) :
— ра — pbE
МС — с + Eo
%3D
E
pb
b) Find the fishing effort that maximizes resource rent, EMey, and the corresponding harvest, HMEY. What happens to EMEY
and HMEY if p increases?
Putting MR(E) equal to MC(E), which is a in this case, gives the following effort level
ра — с
2bp
a
C
MR(E) = pa – 2pbE = MC = c → EMEY
→ if p increases then EMEY increases
2bp
2b
a
C
a
2
→ HMEY= AEMEY – bEmeY
if p increases then HMEY increases
%3D
2b 2bp
2p
Transcribed Image Text:In a fishery the long-run harvest function (harvest volume) is H(E) = aE – bE² a, b positive constants, E is fishing effort. Total cost is TC (E)= cE, with c=unit cost of effort, Total revenue is TR(E)=p*H(E), with p = constant price of fish. a) Find the open-access equilibrium values of effort and harvest, E, and Ho , respectively pH(E) ра — с AR(E) : — ра — pbE МС — с + Eo %3D E pb b) Find the fishing effort that maximizes resource rent, EMey, and the corresponding harvest, HMEY. What happens to EMEY and HMEY if p increases? Putting MR(E) equal to MC(E), which is a in this case, gives the following effort level ра — с 2bp a C MR(E) = pa – 2pbE = MC = c → EMEY → if p increases then EMEY increases 2bp 2b a C a 2 → HMEY= AEMEY – bEmeY if p increases then HMEY increases %3D 2b 2bp 2p
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