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Supply and demand equation

  1. Apr 1, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    Can anyone tell me where I'm wrong?


    2.The attempt at a solution
    supply equation: (0.9-0.6)/(750-450) = 0.001 then went out to find that if price is 0 then the supply is -150... my equation is p=0.001q-150

    demand equation: (0.9-0.6)/(495-645) = -0.002q and if price is 0 then demand goes up to 945... my equation is p= -0.002+945
     
  2. jcsd
  3. Apr 2, 2016 #2

    SteamKing

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    Did you check these equations by substituting the supply and demand quantities in them to see if you obtained the listed prices?

    For example, if the supply equation is p = 0.001q - 150, then when q = 450, then the price p should equal $0.60. Does your equation give that price?
    Same thing for the demand equation.
     
  4. Apr 2, 2016 #3
    i got it wrong after checking
     
  5. Apr 2, 2016 #4

    Simon Bridge

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    Did you figure out which bit of the equation didn't work?
    You could also get a bit of graph paper out, draw axes, plot the points, and draw lines through them: that will help you understand where you made a mistake.

    ie. what is the price for zero bushels?
     
  6. Apr 2, 2016 #5
    i got the answer after using the reciprocals of the initial slope i used. for example instead of (0.9-0.6)/(750-450).. i used (750-450)/(0.9-0.6).... which kind of irritates me because i don't understand the logic behind using that slope over the other
     
  7. Apr 2, 2016 #6

    Ray Vickson

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    If you do them both correctly, they will both ways will yield identical results.

    If you want the equation of a straight line through the tow points ##(x_1,y_1)## and ##(x_2,y_2)##, you either can write it as
    [tex] y-y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x- x_1) \; \Longrightarrow \; y = (y_1 - r x_1) + rx, [/tex]
    where ##r = (y_2 - y_2)/(x_2 - x_1)##, or you else can write it as
    [tex] x - x_1 = \frac{x_2 - x_1}{y_2 - y_1} (y - y_1) \; \Longrightarrow \: x = (x_1 - s y_1) + s y, [/tex]
    where ##s = (x_2 - x_1)/(y_2 - y_1) = 1/r##.
     
  8. May 5, 2016 #7
    Both of your equations are wrong, but you knew that you were suppose to use slope formula to find your m value so those values are correct. Now you must use point-slope formula in order to get the equations. First you should know your coordinates for supply, in this case they are (450, .60) and (750, .90), now you plug in your values into the equation. Since the equation only calls for x subscript 1 and y subscript 1 you will only use point (450, .60) for the formula.
    • This will leave you with the equation for supply looking like: -.60= .001(x-450) so you must move -.60 to the right hand side
    • P= .001(x-450)+.60 is now your new equation but it is still not correct because you must distribute your .001 into (x-450)
    • now you have P= .001x -.45 +.60, after this you will add the values which dont have the variable in them leaving you with the correct equation
    • P= .001x +.15 is now your correct equation
    Lastly, you repeat the same steps to find your demand equation, but you must use the values that pertain to the demand only. This leaves you with different x values but the same y values (which are the .60 and .90).
     
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