hw04a_2008 - UNC-Wilmington Department of Economics and...

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UNC-Wilmington ECN 321 Department of Economics and Finance Dr. Chris Dumas Homework 4 Solutions (1) max U = 10 + 3ln(X) + X -1 X subject to: no constraints F.O.C.: 0 X X 1 3 0 X U 2 = - + = - Solve FOC for X: 2 2 X X 1 3 0 X X 1 3 0 - - = = - + Multiply both sides by X 2 . . . 1 X 3 * = 3 / 1 X * = X * is the solution; that is, X * is the particular value of X that maximizes U. (2) max U = 20 + 100X – 2X 2 X subject to: no constraints F.O.C.: 0 X 4 100 0 X U 1 = - + = Solve FOC for X: 25 4 / 100 X X 4 100 0 X 4 100 0 * = = = = - + X * is the solution; that is, X * is the particular value of X that maximizes U. 1
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UNC-Wilmington ECN 321 Department of Economics and Finance Dr. Chris Dumas (3) max U = 40 + 60X 1 - (1/4)X 1 2 + 100X 2 - (1/3)X 2 2 X 1 , X 2 subject to: no constraints F.O.C.'s: (1) 0 X ) 4 / 1 ( 2 60 X U 1 1 = - = (2) 0 X ) 3 / 1 ( 2 100 X U 2 2 = - = Solve FOC (1) for X 1 : X 1 * = 60/[2(1/4)] = 120 Solve FOC (2) for X 2 : X 2 * = 100/[2(1/3)] = 150 (4) max U = 300X 1 + 6000X 2 - (1/3)X 1 2 - 3X 2 2 – (1/4)X 1 X 2 X 1 , X 2 subject to: no constraints F.O.C.'s: (1) 0 X ) 4 / 1 ( X ) 3 / 1 ( 2 300 X U 2 1 1 = - - = (2) 0 X ) 4 / 1 ( X 6 6000 X U 1 2 2 = - - = The FOC equations provide us with two equations in two unknowns. Solve these two equations for the solution values of X 1 and X 2 . There are several ways to work the algebra to solve these two equations. One way is shown below: Solve FOC (1) for X 2 : 300 - 2(1/3)X 1 – (1/4)X 2 = 0 300 - 2(1/3)X 1 = (1/4)X 2 1200 - 8(1/3)X 1 = X 2 . . . call this equation (3 ) (Note: It’s not a final result for X 2 , because the right-hand-side of the equation still contains variable X 1 .) Substitute equation (3) back into FOC (2) and solve for X 1 : 6000 - 6X 2 – (1/4)X 1 = 0 6000 - 6[1200 - 8(1/3)X 1 ] – (1/4)X 1 = 0 6000- 7200 + 48(1/3)X 1 – (1/4)X 1 = 0 - 1200 + (15.75)X 1 = 0 (15.75)X 1 = 1200 X 1 * = 1200/(15.75)
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hw04a_2008 - UNC-Wilmington Department of Economics and...

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