Question 4 (20 marks) In a fishing village off the Atlantic coast of Norway, the local...

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Question 4 (20 marks) In a fishing village off the Atlantic coast of Norway, the local demand function in (lbs) pounds for wild salmon is given by q(t) = 10t - 2001 +15,000; where 1t12 is the month of the year. The price, in dollars, of a pound of wild salmon also varies during the year according to p(t)=1.51+ a) At what month of the year is the demand the lowest? Justify. (2 marks) b -200 a=10>0 : The parabola is open upwards. The vertex h=- =10 2a 20 is the minimum. The demand is lowest at the 10th month of the year (October) b) Give the monthly revenue function R(t). R(t) = P(1)9(t)=(1.51+) (10r (10t - 200t+15,000) $ (3 marks) dR c) Determine the revenue monthly rate of change R' (t) = Do not simplify. dt (7 marks) R (1) = [(1.57 +23) (101 - 2001 +15,000) = + dt 3)] (107 = 1.5 3 R(t)= 21 (10t - 200t +15,000) + 1)+(1.57+3) | | [(10r - 2001 +15,000)] 101 - 2001 +15,000) + (1.5 7 +27) (201-200)$/month d) Determine the marginal revenue dR 1.5 3 dR = dq dR dq = dR dt dt 21 12 = dt dq dq 1.5 3 21 12 dt = (10t - 2001 +15,000) + (1.51+1 20t-200 (1012- 2-2007 +15,000) + 1.57+ 3 (201-200) dt (1012 2-2001 +15,000) 331 201-2 -200) dR dq I-10 = Clearly, a value 10 will make the denominator zero. Interpret mathematically your result. t=10 is a (singular) critical point of the function R(t) (8 marks Question 4 (20 marks) In a fishing village off the Atlantic coast of Norway, the local demand function in (lbs) pounds for wild salmon is given by q(t) = 10t - 2001 +15,000; where 1t12 is the month of the year. The price, in dollars, of a pound of wild salmon also varies during the year according to p(t)=1.51+ a) At what month of the year is the demand the lowest? Justify. (2 marks) b -200 a=10>0 : The parabola is open upwards. The vertex h=- =10 2a 20 is the minimum. The demand is lowest at the 10th month of the year (October) b) Give the monthly revenue function R(t). R(t) = P(1)9(t)=(1.51+) (10r (10t - 200t+15,000) $ (3 marks) dR c) Determine the revenue monthly rate of change R' (t) = Do not simplify. dt (7 marks) R (1) = [(1.57 +23) (101 - 2001 +15,000) = + dt 3)] (107 = 1.5 3 R(t)= 21 (10t - 200t +15,000) + 1)+(1.57+3) | | [(10r - 2001 +15,000)] 101 - 2001 +15,000) + (1.5 7 +27) (201-200)$/month d) Determine the marginal revenue dR 1.5 3 dR = dq dR dq = dR dt dt 21 12 = dt dq dq 1.5 3 21 12 dt = (10t - 2001 +15,000) + (1.51+1 20t-200 (1012- 2-2007 +15,000) + 1.57+ 3 (201-200) dt (1012 2-2001 +15,000) 331 201-2 -200) dR dq I-10 = Clearly, a value 10 will make the denominator zero. Interpret mathematically your result. t=10 is a (singular) critical point of the function R(t) (8 marks