## PS1_Solutions

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## PS1_Solutions

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- Pages:
- 3
- School:
- University of California, Los Angeles
- Course:
- Psych 10 - Introductory Psychology

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ECON 11 Problem Set 1 Solutions 1 Unconstrained Optimization Single Variable Consider the following function f x 3x3 5x2 x where x 0 2 a Find the x s that are critical value points of f x values of x that are a potential minimum or maximum Are those critical values maxima or minima Hint Check the second order derivative Check also the value of the function at the endpoints 0 and 2 to make sure you have the correct answer Solution Take FOC to get critical points f 0 x 9x2 10x 1 0 x 1 Check SOC 1 9 1 f 00 x 18x 10 f 00 0 f 00 1 0 9 Thus there are two critical points x need to check for the END POINTS 1 9 is a local maximum x 1 is a local minimum But we also f 0 0 f 2 6 Compare with 1 13 f 1 1 f 9 243 Therefore f 1 1 f 2 6 are the global minimum and maximum respectively 2 Unconstrained Optimization Two Variables Consider the function f x1 x2 x13 3x23 9x1 x2 1 a Find a minimum given that x1 x2 1 Solution Take FOC f1 3x12 9x2 0 f2 9x22 9x1 0 We get x1 x2 0 and x1 3 9 x2 3 3 But because x1 x2 1 we only have to consider x1 3 9 x2 3 3 Check SOC f11 6x1 6 3 9 0 f12 9 f12 9 f22 18x2 18 3 3 0 3 3 2 f11 f22 f12 6 9 18 3 9 9 182 92 0 Therefore x1 3 9 x2 3 3 is a minimum Note the followings is not required and will not be tested 2 0 when x x 0 So 0 0 is a saddle point But you First if you are interested f11 f22 f12 1 2 are not required to know what a saddle point is Also this function does not have a global maximum or global minimum For example x1 x2 0 then f x1 x2 x1 x2 0 then f x1 x2 3 Constrained Optimization Julia maximizes the following utility function 2 7 5 7 u x1 x2 x1 x2 subject to the budget constraint p1 x1 p2 x2 I where p1 p2 x1 x2 I 0 a Find the x1 x2 that maximizes u x1 x2 i Step 1 Identify the type of problem This is a CONSTRAINED optimization problem with two variables Therefore we need to apply the Lagrangian multiplier method ii Step 2 Setup the Lagrangian 2 7 5 7 L x1 x2 x1 x2 I p1 x1 p2 x2 iii Step 3 Derive the FOCs and solve them for the critical points 2 5 7 5 7 2 5 7 5 7 L1 x1 x2 p1 0 x1 x2 p1 7 7 5 2 7 2 7 5 2 7 2 7 L2 x1 x2 p2 0 x1 x2 p2 7 7 L3 I p1 x1 p2 x2 0 p1 x1 p2 x2 I From the above two equations we get 1 2x2 p1 5x1 p1 x2 2 5x1 p2 2p2 2 1 2 3 substituting into equation 3 we get x1 2I 7p1 x2 5I 7p2 iv The SOCs for a constrained optimization problem are not required and will not be tested so I will omit them here but the critical point is a maximizer b Show that the maximizer x2 p1 p2 I is decreasing in p2 and increasing in I Hint Use the partial derivatives This question asks to compute the change in x2 as p2 and I increase Since the change in the function x2 with respect to one of the variables is determined by the partial derivative this question is essentially asking you to compute the derivatives and to check their sign x2 5I 2 0 p2 7p2 x2 5 0 I 7p2 c Does x2 p1 p2 I change if p1 changes Hint partial derivatives x2 0 p1 Therefore x2 p1 p2 I does not change if p1 changes 3

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