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  • Sep 1, 2010, 10:28 PM
    asmith_85
    can someone solve this please...
    The following represents demand for widgets: QD = 100 – 5P +0.004M – 5PR, where P is the price of widgets, M is income, and PR is the price of a related good, the wodget. Supply of widgets is determined by QS = 150 + 5P.
    a) Determine whether widgets are a normal or inferior good, and whether widgets and wodgets are substitutes or complements.
    b) Assume that in 2009 M = $50,000 and PR = $10. Solve algebraically to determine the 2009 equilibrium price and quantity of widgets.
    c) Generate a 2009 supply/demand graph in Excel. Be sure that P is the vertical axis and Q the horizontal. Does the graphical equilibrium correspond to your algebraic equilibrium?
    d) Now assume two events occur in 2010: the wodget price drops to $8 and supply conditions change such that QS = 120 + 7P. Solve algebraically for the new 2010 equilibrium price and quantity of widgets after these two changes
  • Sep 2, 2010, 04:08 AM
    ArcSine
    a) Remember that a normal good is one for which the demand increases when consumer income increases. An inferior good, OTOH, sees its demand increase when consumer income decreases---consumers have to give up buying their preferred good, and switch to the cheaper (inferior) good. To make that call, look at the coefficient of M (income) in your demand equation: is it positive or negative? In other words, will widget demand Qd increase or decrease as a result of an increase in income?

    On a similar note, review the definitions for complements and substitutes, then examine the sign of the coefficient for wodgets' price PR, to determine the direction in which widget demand Qd will move, given an increase of wodget price PR.

    b) You're given assumed values for M and PR. This reduces your demand equation Qd into a simple linear with one variable, price P. Now you've got Qs and Qd in a 'two-equation, two-unknown' setup. Find the solution (equilibrium) algebraically. You might want to review your algebra text for solving via "elimination" or by "substitution"; either method will get the job done quickly and efficiently.

    d) This is just re-do of (b), after making those two changes. Note that those are the only two changes; hence, assume M remains at 50K, as in (b).

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