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MMEprofitmaxsoln

# MMEprofitmaxsoln - 2011 Steven Tschantz Profit maximization...

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© 2011 - Steven Tschantz Profit maximization Steven Tschantz 2/1/11 Exercises Complete the following exercises. You can save a copy of this notebook, delete the material above, insert and evaluate your commands computing answers below, save your results, and email your final version to [email protected] 1. Find the coefficients in a linear demand function assuming demand is 10000 at a price of \$50 and 8000 at a price of \$55. Define the form of the demand function. We clear the symbols we will use for the unknown coefficients to make sure they are undefined variables. In[1]:= Clear @ qintercept,qslope D ; quantity @ p_ D : = qintercept + qslope * p; Write the conditions to be satisfied as a list of equations. In[3]:= quantityconditions = 8 quantity @ 50. D 10000.,quantity @ 55. D 8000. < Out[3]= 8 qintercept + 50.qslope 10000.,qintercept + 55.qslope 8000. < Solve for the unknown coefficients. In[4]:= quantitysolns = Solve @ quantityconditions, 8 qintercept,qslope <D Out[4]= 88 qintercept fi 30000.,qslope fi - 400. << Simultaneously assign the solution values to the corresponding symbols. Note that this makes qintercept and qslope numerical values so we can no longer use these as variable symbols without clearing. In[5]:= 8 qintercept,qslope < = 8 qintercept,qslope < .quantitysolns @@ 1 DD Out[5]= 8 30000., - 400. < What is the maximum price giving non-negative demand? In[6]:= pmaxsoln = Solve @ quantity @ p D 0,p D Out[6]= 88 p fi 75. << In[7]:= pmax = p .pmaxsoln @@ 1 DD Out[7]= 75. So here is a plot of the demand function.

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In[8]:= Plot @ quantity @ p D , 8 p,0,pmax <D Out[8]= 10 20 30 40 50 60 70 5000 10000 15000 20000 25000 30000 2. Solve for the profit maximizing price for this demand, assuming a constant marginal cost of \$25. In[9]:= Clear @ fixedcost,marginalcost D ; cost @ q_ D : = fixedcost + marginalcost * q; In[11]:= profit @ p_ D : = p * quantity @ p D - cost @ quantity @ p DD ; In[12]:= profit @ p D Out[12]= - fixedcost - marginalcost H 30000. - 400.p L + H 30000. - 400.p L p In[13]:= marginalcost = 25.; In[14]:= profit @ p D Out[14]= - fixedcost - 25. H 30000. - 400.p L + H 30000. - 400.p L p We can't plot this without a value for fixedcost. Adding fixedcost shifts the profit curve down but doesn't change where the maximum occurs. Here is a way of temporarily assigning fixedcost equal to 0 so we can see the profit. In[15]:= Plot @ profit @ p D . 8 fixedcost fi 0. < , 8 p,0,pmax <D Out[15]= 10 20 30 40 50 60 70 - 600000 - 400000 - 200000 200000 In[16]:= Solve @ profit @ p D 0,p D Out[16]= :: p fi 0.00125 K 40000. - 40. 250000. - 1.fixedcost O> , : p fi 0.00125 K 40000. + 40. 250000. - 1.fixedcost O>> 2 MMEprofitmaxsoln.nb
In[17]:= Solve @H profit @ p D . 8 fixedcost fi 0. <L 0,p D Out[17]= 88 p fi 25. < , 8 p fi 75. << Now to find the optimum price. In[18]:= firstordercondition = profit' @ p D 0 Out[18]= 40000. - 800.p 0 And the fixedcost parameter has indeed disappeared. In[19]:= profmaxsoln = Solve @ firstordercondition,p D Out[19]= 88 p fi 50. << In[20]:= profmaxprice = p .profmaxsoln @@ 1 DD Out[20]= 50. Let's go ahead and just assume fixed cost is zero. In[21]:= fixedcost = 0.; In[22]:= quantity @ profmaxprice D Out[22]= 10000. In[23]:= profmaxprice * quantity @ profmaxprice D Out[23]= 500000. In[24]:= cost @ quantity @ profmaxprice DD Out[24]= 250000.

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