SOLUTION: At $0.60 per bushel, the daily supply for wheat is 450 bushels and the daily demand is 645 bushels. When the price is raised to $0.90 per bushel, the daily supply increases to 750

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: At $0.60 per bushel, the daily supply for wheat is 450 bushels and the daily demand is 645 bushels. When the price is raised to $0.90 per bushel, the daily supply increases to 750       Log On


   



Question 799124: At $0.60 per bushel, the daily supply for wheat is 450 bushels and the daily demand is 645 bushels. When the price is raised to $0.90 per bushel, the daily supply increases to 750 bushels and the daily demand decreases to 495 bushels. Assume that the supply and demand equations are linear.
(A) Find the supply equation.
(B) Find the demand equation.
(C) Find the equilibrium price and quantity.

Answer by stanbon(75887) About Me  (Show Source):
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At $0.60 per bushel, the daily supply for wheat is 450 bushels and the daily demand is 645 bushels. When the price is raised to $0.90 per bushel, the daily supply increases to 750 bushels and the daily demand decreases to 495 bushels. Assume that the supply and demand equations are linear.
(A) Find the supply equation.
(0.6,450) and (0.9,750)
slope = (750-450)/(0.9-0.6) = 300/0.3 = 1000
450 = 1000*0.6+b
450 = 600 + b
b = -150
S(x) = 1000x - 150
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(B) Find the demand equation.
(0.6,645) and (0.9,495)
slope = -150
645 = -150*0.6 + b
b = 555
D(x) = -150x + 555
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(C) Find the equilibrium price and quantity.
Solve S(x) = D(x)
1000x - 150 = -150x + 555
1150x = 705
x = 0.613 (price)
S(0.613) = 1000*0.613 - 150 = 613-150 = 463 (supply)
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Cheers,
Stan H.
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