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172A-PS1

# 172A-PS1 - Econ172A #1 Winter2010 1 Consider the linear...

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Econ 172A Solutions to Problem Set #1 Winter 2010 1. Consider the linear programming problem ( ) , 0 max , 2 3 s.t. 4 x y f x y x y x y M = + + a. Solve this problem graphically when M = 8. This part is illustrated using the black constraint above. The constraint can be re-written as 1 2 4 y x = − + which has a slope of –1/4. A level curve of f can be written as 2 3 3 c y x = − + which has a slope of –2/3. Since the level curve is steeper than the constraint (and the objective function is increasing in both x and y ), the solution occurs at the x -intercept. x * = 8, y * = 0, f ( x *, y *) = 2(8) + 3(0) = 16 b. Solve this problem graphically (and intuitively) for all M > 0. This part is partially illustrated using the red constraint above (for M = 12). M does not change the slope of the constraint (or of a level curve of f ). Regardless of M , the solution will occur at the x -intercept. x * = M , y * = 0, f ( x *, y *) = 2(M) + 3(0) = 2 M c. Comment how the corner of the solution changes as M changes. Briefly explain why. The corner of the solution won’t change because M does not affect the slope of the constraint or of a level curve of f . x 2 y 8 12 3

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2. Consider the linear programming problem ( ) , 0 max , 2 3 s.t. 4 5 x y f x y x y x y M x y = + + + a. Solve this problem graphically when M = 8. The first (black) constraint can be re-written as 1 2 4 y x = − + which has a slope of –1/4. The second (red) constraint can be re-written as 5 y x = − + which has a slope of –1. A level curve of f can be written as 2 3 3 c y x = − + which has a slope of –2/3. Since the level curve is steeper than the constraint and flatter than the second constraint, the solution occurs at their intersection. ( ) ( ) ( ) 4 8 3 3 5 * 1, * 5 1 4, *, * 2 4 3 1 11 x y y x y y x f x y + = = + = = = = = + = (To get the intuition it’s important to use the graph.) x 2 y 5 8 5
b. Solve this problem graphically for all M > 0. There are three types of solutions. When M 20 (illustrated in green), the first constraint is redundant. This means whenever the second constraint is satisfied the first is guaranteed to be satisfied. The feasible region is just the area on or below the second (solid black) constraint. Since the level curves are flatter than the second constraint the solution occurs at the y -intercept of the second constraint. x * = 0, y * = 5, f ( x *, y *) = 2(0) + 3(5) =15. When M 5 (illustrated in green), the second constraint is redundant. x 2 y 5 8 5 x 2 y 5 8 5

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The feasible region is just the area on or below the first (green) constraint. Since the level curves are steeper than the first constraint the solution occurs at the x - intercept of the first constraint. x * = M , y * = 0, f ( x *, y *) = 2( M ) + 3(0) = 2 M . When 5 < M < 20, we’re in an in-between case. The solution looks like the one in part (a). The intersection of the two constraints is part of the feasible region. Since the level curve is steeper than the constraint and flatter than the second constraint, the solution occurs at their intersection.
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172A-PS1 - Econ172A #1 Winter2010 1 Consider the linear...

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