Problem Set1 Solutions

Problem Set1 Solutions - BPUB250 Problem Set1 Due: January...

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BPUB250 Problem Set1 Due: January 25-26, 2012 in Class Question 1 Monica wants to solve the following problem: Max B 2 SN s.t. B+S+N=24 Note that N=24-B-S, so we can use the substitution method and rewrite it as: Max B 2 S(24-B-S) The first order conditions for this unconstrained maximization problem are: FOC S : 24B 2 -B 3 -2SB 2 = 0 (1) FOC B : 48BS-3B 2 S-2BS 2 =0 (2) Rewrite equation (1) as : B 3 = 24B 2 -2SB 2 B = 24-2S (1-1) Substitute it in equation (2) to obtain: 2BS 2 = 48BS-3B 2 S 2S = 48-3B = 48-3(24-2S) 4S = 24 S * =6 Substitute back in (1-1) to obtain, B * =24-2(6)=12 Because B+S+N=24, we must have N * =6. Hence the solution is : B * =12, S * =6, N * =6 The maximum utility that can be obtain is U=12 2 *6*6=5184 Note1 : We can solve this maximization problem with Lagrangian Method and still get the same answer! Note2 : Substitution method will not work well if the problem becomes more complicated with 3+ variables. In that case, we will need to use Lagrangian Method. Note3 : In BPUB250, we will mostly focus on 2 variable cases. Even when you face 3 variables, you will be able to solve them with substitution method. (or MRS = - Price Ratio)
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<Lagrangian Method> £=ܤ ܵܰ + ߣ(24−ܤ−ܵ−ܰ) ߲£ ߲ܤ = 2ܤܵܰ − ߣ = 0 (݅) ߲£ ߲ܵ ܰ − ߣ = 0 (݅݅) ߲£ ߲ܰ ܵ − ߣ = 0 (݅݅݅) ߲£ ߲ߣ = 24 − ܤ − ܵ − ܰ = 0
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This note was uploaded on 04/04/2012 for the course BPUB 250 taught by Professor Seim during the Spring '08 term at UPenn.

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Problem Set1 Solutions - BPUB250 Problem Set1 Due: January...

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