ECON 110 Fall 2007 Problem Set 1 Solutions

# ECON 110 Fall 2007 Problem Set 1 Solutions - Econ 110 PS1...

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Econ 110 PS1 : Solutions By Omair Sarwar 1.1 (d) is not a function as y = + x is not a one to one mapping. 1.2 f(3) = 3 2 + 6 + 7 = 22 1.3 Q=10 => C = 100 + 25(10) = 350 1.4 U shaped parabola with y intercept (0,5) 1.5 1.6 Q= 100 – 5p => 5p = 100 – Q => p = 20 – 1/5Q, draw graph 1.7 m = (1-3)/(5-1) = -1/2 1.8 TC = 1000 + 20Q 2.1 2x + y = 4 x – y = 1 Adding both gives 3x = 5 => x= 5/3 ; y = 2/3 2.2 Q d = 100 – 5p Q s = 50 + 5p Note that at intersection Q d = Q s => 100 – 5p = 50 + 5p => p = 10 2.3 2.1 was solved by elimination above and 2.2 was solved using substitution. 3.1 gradient fn = f’(x) = -1/x 2 at x= 2 slope is -1/4 ; x=5 slope is -1/25

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Unformatted text preview: 3.2 MC = dC/dQ = 5 + 4Q. At Q = 1, MC = 9 ; Q = 5 , MC = 25 3.3 (1) 2x + 2 ; (2) 15x 4 + 8x + 10 ; (3) -1/x 2 + 1/2(x-1/2 ) 4.1 (1) Trivial solution, line x + z = 1 (2) Graph is L shaped curve with minimum point at (1,1) (3) Graph is inverse(mirror image) of that for min fn. with maximum pt at (1,1) (4) Level curve is hyperbola with z : Æ ∞ , -∞ as x : Æ 0 4.2 fn : U = l + r , for U = 10 level curve is a straight line l + r = 10 4.3 ∂ F/ ∂ x = 2x + 2z ∂ F/ ∂ z = 2x + 30z 2 5.1 For x to be a maximum or minimum f’(x) = 0 =&gt; 50 – 4x = 0 =&gt; x = 25/2...
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• Fall '08
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• UCI race classifications, Tour de Georgia, 1.7 M, Omair Sarwar

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