problem set 1_ans_econ2302

# Principles of Microeconomics (7th Edition) (Case/Fair Economics 7e Series)

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Economics 2302 Principles of Microeconomics Isaac McFarlin University of Texas at Dallas Solutions to Problem Set 1 1. 2. Two linear relationships in two unknowns: X + Y = 100 and Y – X = 20. The latter implies Y = 20 + X, so sub into the former to get X + (20 + X) = 100. Therefore, 2X = 80, or X = 40. Compute Y = 100 – X = 100 – 40 = 60. y x Y = X + 20 (slope = 1) 100 100 Y = 100 – X (slope = -1) 60 40 20 y 15 x 5 Slope = -3

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3. Equate price (P) in the two relationships: 100 – Q = 1800 / Q + 10 100 Q – Q 2 = 1800 + 10Q (#) Q 2 – 90Q + 1800 = 0 (*) (Q – 60)(Q – 30) = 0 One way to solve this is to note that the quadratic is a complete square and can be factored as in (*). You’ll notice that Q = 30 or Q = 60 lets the quadratic equal zero. Another way to solve the problem is by using the quadratic formula. Suppose that a
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Unformatted text preview: quadratic takes the following form: (1) 2 = + + c bQ aQ then you can use the following formula to uncover the roots or the Q values that allow the equation to equal zero. (2) a ac b b Q Q 2 4 , 2 2 1 − ± − = Looking at equation (#) in this case, we find that a = 1, b = -90, and c = 1800 and, therefore, 1 Q =30 and 2 Q = 60. The next step is to solve for P. We can find this by substituting both values of Q into one of the original equations. For Q = 30, P = 70 and for Q = 60, P = 40. Note that both sets of prices (P) and quantities (Q) are positive. These two solutions correspond to the intersections on the graph below. P Q 70 100 100 30 40 60 P = 1800/Q + 10 P =100 - Q...
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• Spring '06
• McFarlin

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