8. Optimization with inequality constraints. Find
maxf (x,y) = xy s.to.x+y2 ≤2
Approach this formally via Lagrangean. Write all the Karush-Kuhn-Tucker con- ditions and solve the resulting system of equations.
We have 3 inequality constraints here, hence three Lagrange multipliers. The full Lagrangean and the FOCs wrt to (x, y) is
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$ = xy−λ1 x+y2 −2 −λ2(−x)−λ3(−y)
And three complementary slackness conditions
λ1 x+y2−2 = 0 λ2x = 0 λ3y = 0
In principle we need to solve the FOCs and the slackness conditions as a system of equations in (x, y, λ1, λ2, λ3) . That will deliver (maybe a couple) candidates for the optimum. This is a hard math exercise. We will take an easier path.
Eyeball the constraints. They imply that the area where the optimum can be is under the line x + y2 = 2 and the solution most likely satisfies x ≥ 0,y ≥ 0. We can ignore these constraints for now and check them later. This implies that λ2 = λ3 = 0. We also see that x + y2 ≤ 2 constraint will be satisifed with equality. This implies thatλ1 =0.Callλ=λ1.
Then the simplified Lagrangean is
L=xy−λ x+y2 −2 . The first order conditions
= y−λ1+λ2=0
= x−2yλ1+λ3=0
= x+y2−2=0. 4
Solve the system, i.e. λ = y, then x = 2y2 and 3y2 = 2 and obtain the answers x = 34 , y = 23 , λ = 23 .
These x and y satisfy the omitted constraints, hence we were correct to omit them. Further, since this solution to the problem with fewer contraints satisfies all the constraints in the original promlem, this solution is the optimum in the original problem as well.
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