1. A Consumer has a utility function given by U=2xy income is 120 with Px=birr4 and Py= a/ Calculate optimum points of X and Y that gives the consumer high satisfaction? (2pts) b/ What is the value of MRSxy at equilibrium? (2pts) 2. Suppose the short run cost function of a firm is given by: TC=Q^3-6Q^2+15Q+20 A. Find the expression of TFC & TVC (2pts) B. Derive the expressions of AFC, AVC, AC and MC. (4pts) C. Find the levels of output that minimize MC and AVC and then find the minimum vaiues of MC and AVC. (4pts)

Let's solve each part of the questions step by step.<br /><br />### Question 1<br /><br />#### Part (a)<br />The consumer's utility function is given by \( U = 2xy \). The consumer has an income of \$120, with the price of good X (\(P_x\)) being birr 4 and the price of good Y (\(P_y\)) being unknown. <br /><br />To find the optimum points of X and Y that give the consumer high satisfaction, we need to maximize the utility function subject to the budget constraint. The budget constraint is given by:<br />\[ P_x x + P_y y = I \]<br />where \( I \) is the income.<br /><br />Given:<br />\[ P_x = 4 \]<br />\[ I = 120 \]<br /><br />So the budget constraint becomes:<br />\[ 4x + P_y y = 120 \]<br /><br />To maximize utility, we use the method of Lagrange multipliers. The Lagrangian function is:<br />\[ \mathcal{L} = 2xy + \lambda (120 - 4x - P_y y) \]<br /><br />Taking partial derivatives and setting them to zero:<br />\[ \frac{\partial \mathcal{L}}{\partial x} = 2y - 4\lambda = 0 \]<br />\[ \frac{\partial \mathcal{L}}{\partial y} = 2x - P_y \lambda = 0 \]<br />\[ \frac{\partial \mathcal{L}}{\partial \lambda} = 120 - 4x - P_y y = 0 \]<br /><br />From the first equation:<br />\[ 2y = 4\lambda \]<br />\[ \lambda = \frac{y}{2} \]<br /><br />From the second equation:<br />\[ 2x = P_y \lambda \]<br />\[ 2x = P_y \left(\frac{y}{2}\right) \]<br />\[ 4x = P_y y \]<br />\[ x = \frac{P_y y}{4} \]<br /><br />Substitute \( x = \frac{P_y y}{4} \) into the budget constraint:<br />\[ 4\left(\frac{P_y y}{4}\right) + P_y y = 120 \]<br />\[ P_y y + P_y y = 120 \]<br />\[ 2P_y y = 120 \]<br />\[ P_y y = 60 \]<br />\[ y = \frac{60}{P_y} \]<br /><br />Now substitute \( y = \frac{60}{P_y} \) back into \( x = \frac{P_y y}{4} \):<br />\[ x = \frac{P_y \left(\frac{60}{P_y}\right)}{4} \]<br />\[ x = \frac{60}{4} \]<br />\[ x = 15 \]<br /><br />So the optimal points are:<br />\[ x = 15 \]<br />\[ y = \frac{60}{P_y} \]<br /><br />#### Part (b)<br />The Marginal Rate of Substitution (MRS) is given by the ratio of the marginal utilities of the two goods:<br />\[ MRS_{xy} = \frac{MU_x}{MU_y} \]<br /><br />The marginal utilities are:<br />\[ MU_x = \frac{\partial U}{\partial x} = 2y \]<br />\[ MU_y = \frac{\partial U}{\partial y} = 2x \]<br /><br />At equilibrium, \( MRS_{xy} = \frac{P_x}{P_y} \):<br />\[ \frac{2y}{2x} = \frac{P_x}{P_y} \]<br />\[ \frac{y}{x} = \frac{P_x}{P_y} \]<br /><br />Using the optimal values \( x = 15 \) and \( y = \frac{60}{P_y} \):<br />\[ \frac{\frac{60}{P_y}}{15} = \frac{4}{P_y} \]<br />\[ \frac{4}{P_y} = \frac{4}{P_y} \]<br /><br />Thus, the value of \( MRS_{xy} \) at equilibrium is:<br />\[ MRS_{xy} = \frac{P_x}{P_y} = \frac{4}{P_y} \]<br /><br />### Question 2<br /><br />#### Part (A)<br />The total cost (TC) function is given by:<br />\[ TC = Q^3 - 6Q^2 + 15Q + 20 \]<br /><br />Total Fixed Cost (TFC) is the cost that does not vary with output (Q), which is the constant term in the TC function:<br />\[ TFC = 20 \]<br /><br />Total Variable Cost (TVC) is the part of the total cost that varies with output:<br />\[ TVC = TC - TFC \]<br />\[ TVC = Q^3 - 6Q^2 + 15Q \]<br /><br />#### Part (B)<br />Average Fixed Cost (AFC) is given by:<br />\[ AFC = \frac{TFC}{Q} = \frac{20}{Q} \]<br /><br />Average Variable Cost (AVC) is given by:<br />\[ AVC = \frac{TVC}{Q} = \frac{Q^3 - 6Q^2 + 15Q}{Q} = Q^2 - 6Q + 15 \]<br /><br />Average Cost (AC) is given by:<br />\[ AC = \frac{TC}{Q} = \frac{Q^3 - 6Q^2 + 15Q + 20}{Q} = Q^2 - 6Q + 15 + \frac{20}{Q} \]<br /><br />Marginal Cost (MC) is the derivative of the total cost function with respect to Q:<br />\[ MC = \frac{d(TC)}{dQ} = \frac{d(Q^3 - 6Q^2 + 15Q + 20)}{dQ} = 3Q^2 - 12Q + 15 \]<br /><br />#### Part (C)<br />To find the levels of output that minimize MC and AVC, we take the first derivative of each with respect to Q and set it to zero.<br /><br />For MC:<br />\[ \frac{d(MC)}{dQ} = \frac{d(3Q^2 - 12Q + 15)}{dQ} = 6Q - 12 \]<br />Set the derivative equal to zero:<br />\[ 6Q - 12 = 0 \]<br />\[ Q = 2 \]<br /><br />For AVC:<br />\[ \frac{d(AVC)}{dQ} = \frac{d(Q^2 - 6Q + 15)}{dQ} = 2Q - 6 \]<br />Set the derivative equal to zero:<br />\[ 2Q - 6 = 0 \]<br />\[ Q = 3 \]<br /><br />Now, we find the minimum values of MC and AVC by substituting these Q values back into their respective functions.<br /><br />Minimum MC at \( Q = 2 \):<br />\[ MC = 3(2)^2 - 12(2) + 15 = 3(4) - 24 + 15 = 12 - 24 + 15 = 3 \]<br /><br />Minimum AVC at \( Q = 3 \):<br />\[ AVC = (3)^2 - 6(3) + 15 = 9 - 18 + 15 = 6 \]<br /><br />So, the levels of output that minimize MC and AVC are \( Q = 2 \) and \( Q = 3 \) respectively, with minimum values of MC and AVC being 3 and 6 respectively.