A Karush-Kuhn-Tucker Example
It’s only for very simple problems that we can use the Karush-Kuhn-Tucker conditions
to solve a nonlinear programming problem. Consider the following problem:
maximize f(x, y) = xy
subject to x+y2≤2
x, y ≥0
Note that the feasible region is bounded, so a global maximum must exist: a continuous
function on a closed and bounded set has a maximum there.
We write the constraints as g1(x, y) = x+y2≤2, g2(x, y ) = −x≤0, g3(x, y) = −y≤
0. Thus the KKT conditions can be written as
y−λ1+λ2= 0
x−2yλ1+λ3= 0
λ1(2 −x−y2) = 0
λ2x= 0
λ3y= 0
x+y2≤2
x, y, λ1, λ2, λ3≥0
In each of the “complementary slackness” equations λi(bi−gi(x1, . . ., xn)) = 0, at
least one of the two factors must be 0. With nsuch conditions, there would potentially
be 2npossible cases to consider. However, with some thought we might be able to reduce
that considerably.
Case 1: Suppose λ1= 0. Then the first KKT condition says y+λ2= 0 and the second
says x+λ3= 0. Since each term is nonnegative, the only way that can happen is if
x=y=λ2=λ3= 0. Indeed, the KKT conditions are satisfied when x=y=λ1=
λ2=λ3= 0 (although clearly this is not a local maximum since f(0,0) = 0 while
f(x, y)>0 at points in the interior of the feasible region).
Case 2: Suppose x+y2= 2. Now at least one of x= 2 −y2and ymust be positive.
Case 2a: Suppose x > 0. Then λ2= 0. The first KKT condition says λ1=y. The second
KKT condition then says x−2yλ1+λ3= 2 −3y2+λ3= 0, so 3y2= 2 + λ3>0,
and λ3= 0. Thus y=p2/3, and x= 2 −2/3 = 4/3. Again all the KKT
conditions are satisfied.
Case 2b: Suppose x= 0, i.e. y=√2. Since y > 0 we have λ3= 0. From the second KKT
condition we must have λ1= 0. But that takes us back to Case 1.
We conclude there are only two candidates for a local max: (0,0) and (4/3,p2/3).
The global maximum is at (4/3,p2/3).