Newbie with simple questions (take 2)

*To*: mathgroup at smc.vnet.net*Subject*: [mg62322] Newbie with simple questions (take 2)*From*: misha <iamisha1 at comcast.net>*Date*: Sat, 19 Nov 2005 23:19:16 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

I am a new user (errr..purchaser) of Mathematica, but I have not been able to find answers to these (probably) simple questions with Mathematica?s help browser. I am trying to use Mathematica to solve a simple system of simultaneous equations. I suppose I could use it to solve the first order conditions (FOCs), but I?m having enough problems as it is. I have more ambitious goals than this, but I thought this would be an easy place to start. By the way, can anyone recommend a book heavy in examples for a beginning user such as myself? Here is the complete problem: P(Q) = a - Q (inverse demand curve) Q = q1 + q2 (Cournot) duopoly C1(q1) = c*q1 (firm 1?s commonly known cost function, with constant marginal cost, c) C2 = cL*q2 with probability t cH*q2 with probability 1 - t (firm 2?s cost functions for constant marginal costs cL < cH, known to firm 2 but unknown with certainty to firm 1) If Firm 2 has constant marginal cost cH, firm 2 chooses q2 to solve max{[(a - q1* - q2) - cH]*q2} If Firm 2 has constant marginal cost cL, then firm 2 chooses q2 to solve max{[(a - q1* - q2) - cL]*q2} The resulting FOCs are: <<the asterisk denotes ?optimal? and the cH in parentheses denotes that it is a function of cH>> q2*(cH) = (1/2)*(a - q1* - cH) q2*(cL) = (1/2)*(a - q1* - cL) Similarly, Firm 1 chooses q1 to solve max{t[(a - q1 - q2*(cH)) - c]*q1 + (1 - t)[(a - q1 - q2*(cL)) - c]*q1}, which yields FOC: q1* = (1/2)*[t(a - q2*(cH) - c) + (1 - t)(a - q2*(cL) - c)] So I want to use Mathematica to do the tedious algebra to get me the following: q2*(cH) = (a - 2cH + c)/3 + (1 - t)(cH - cL)/6 q2*(cL) = (a - 2cL + c)/3 - t(cH - cL)/6 q1* = (a - 2c + tcH + (1 - t)cL)/3