Compilation - Reyes PSET

Compilation - Reyes PSET - Number 1 Setting up the profit...

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Number 1 Setting up the profit function for the firm. Profit is equal to Total Revenue - Total Cost Using the given our profit function ± is : ± = H 4350 - 13 * q L q - H q^2 - 5.5 * q + 150 L q H 4350 - 13 q L q - q I 150 - 5.5 q + q 2 M Solving for First Order Condition, Solve @ D @ ± , q D ± 0, q D Derivative of ± with respect to q gives us : 88 q fi - 40. < , 8 q 35. << The value that we consider will be the positive value, since there is no negative quantity. Therefore, q = 35. Checking for second order condition : Solve @ D @ ± , q, q D ± 0, q D Secind derivative of ± with respect to q gives us : 88 q fi - 2.5 << Which is a negative value, therefore we satisfy the condition for maximization. Solving for maximum profit, we plug in the value of q to the profit function. ± = H 4350 - 13 * 35 L 35 - H 35^2 - 5.5 * 35 + 150 L 35 94 937.5 We get the maximium profit, given the maximum values of q. Number 2 9.4 ² 6 Setting up our production, revenue, cost and profit fucntions : In[122]:= Clear @ r, c D In[129]:= Q = L Out[129]= L In[130]:= R = P 0 * Q Out[130]= L P 0 In[131]:= Cost = W 0 * L + f Out[131]= f + L W 0 In[132]:= Profit = R - Cost Out[132]= - f + L P 0 - L W 0 Solving for FOC, we get the derivative of profit wrt to L :
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In[133]:= D @ Profit, L D Out[133]= P 0 - W 0 The maximum profit can be derived from subtarcting the price with the cost of labor, which is wage. H C L If the second partial of the profit function with respect to L is negative , exhibitng diminishing MPL then we can ensure that our profit is maximized by our Labor. Number 3 Given our inverse demand function and total cost function, we can derive the maximizing condition for a Monopoly. To get TR we multiplythe inverse demand funcion p by q, giving us :
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Compilation - Reyes PSET - Number 1 Setting up the profit...

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