Given r=0,025 k = 1000 tonnes q= 0,05 p = 1 a = 10 Maximizing harvest...

Transcribed Image Text:

Given r=0,025 k = 1000 tonnes q= 0,05 p = 1 a = 10 Maximizing harvest occurs at the MSY The long-run harvest function: H(E)=qEK -K qKE2 The MSY criterion: 2q2K -> Maximize H(E) H'(E) = 0 qK- E=0 E r E = 24 EMSY 2q (2 vessels fishing whole year and 1 vessel fishing a half of year) (Alternatively Maximize TR(E)MR(E)=0+ pqK-2pqK) 2.5 vessel year; H(EMSY) = 62.5 tonnes; TR(EMSY) = 62500$; TC(EMSY) = 25000$ KE = 0 E = 1/4) Show steps how to get the results H (Emsy) = 62,5 tonnes, and how TR is = 62500 $ and TC = 25000 Given r=0,025 k = 1000 tonnes q= 0,05 p = 1 a = 10 Maximizing harvest occurs at the MSY The long-run harvest function: H(E)=qEK -K qKE2 The MSY criterion: 2q2K -> Maximize H(E) H'(E) = 0 qK- E=0 E r E = 24 EMSY 2q (2 vessels fishing whole year and 1 vessel fishing a half of year) (Alternatively Maximize TR(E)MR(E)=0+ pqK-2pqK) 2.5 vessel year; H(EMSY) = 62.5 tonnes; TR(EMSY) = 62500$; TC(EMSY) = 25000$ KE = 0 E = 1/4) Show steps how to get the results H (Emsy) = 62,5 tonnes, and how TR is = 62500 $ and TC = 25000