# The foc and the complementary slackness conditions

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The FOC and the complementary slackness conditions (KT1 and KT2) altogether are knownas the Kuhn - Tucker conditions (KT), from the names of two mathematicians that studiedtheir properties.ExampleWe describe the consumption problem using the Kuhn – Tucker conditions: since we alreadyhave some economic intuition on what the solution must look like, we can better see what theconditions do, and how to solve them. Consider (H) with one physical commodity,1SX+=,f(x) = U(x0,…, xS), g = (g0, g1,…, gS+1) where g0(x0,…, xS) = p w– p0x0– … – pSxSis thebudget constraint, g1(x0,…, xS) = x0, g2(x0,…, xS) = x1… through gS+1(x0,…, xS) = xSare thenon-negativity constraints, x ≥ 0.We construct the Lagrangean:L(x0,…, xS, λ0,…, λS+1) = U(x0,…, xS) + λ0(p w– p0x0– … – pSxS) + Σj>0λjxj-1.We compute the FOCs:∂L/∂ x0: ∂ U(x0,…, xS) /∂ x0λ0p0+ λ1= 0∂L/∂ xS: ∂ U(x0,…, xS) /∂ xSλ0pS+ λS+1= 0We then write the complementary slackness conditions:min(λ0, p w– p0x0– … – pSxS) = 0min(λ1, x0) = 0min(λS+1, xS) = 0Once we set up the system of equations, we use it to study properties of optimal solution tothe programming problem, see how they change when we modify parameters of the problem(such as endowments, income and prices), and at times we use them to solve explicitly for asolution (with many variables, you need a computer to do so).For a simple numerical example, let S = 2 and U(x0, x1, x2) = x0+ (1/2)ln x1+ (1/2)ln x2, while p1= p2= 1, and p0= ½.Also assume that w0= 0, and w1= 6, w2= 4. Then the KT conditions become (herewe use λ0= λ and λj= μj-1for j > 0):
1 –½ λ + μ0= 01/2x1λ + μ1= 01/2x2λ + μ2= 0min(λ, 10– ½ x0– x1– x2) = 0min(μ0, x0) = 0min(μ1, x1) = 0min(μ2, x2) = 0Since 1/2x→ +∞ as x→ 0, the solutionto this system must have x1> 0 and x2> 0. Then, from the minequations, we have μ1= μ2= 0. Since 1/2x > 0 for x > 0, λ > 0 and ½ x0+ x1+ x2= 10. From thesecond and third equation, x1= x2= x. We have a subsystem of four equations and four unknowns, x0,x, λ and μ0:1 –½ λ + μ0= 01/2xλ = 0½ x0+ 2x= 10min(μ0, x0) = 0Suppose that μ0> 0. Then x0= 0, and x = 5, λ = 1/10, so that μ0+ 1 – 1/20 = 0 has no solution. Then,μ0= 0, λ = 2, x = ¼ and x0= 19. Not surprisingly, since prices tomorrow in today’s terms are higher,lots of consumption of the physical good occurs today, and because of risk aversion, the individual(fully) insures himself.A (bit of) geometryThe intuition for the recipe is most easily illustrated for the case when n = 1 (a one-dimensional maximization problem) and there are two constraints, or m = 2.Also assume that:-f, g1and g2are continuously differentiable functions;-the set satisfying the two constraints is the finite union of disjoint intervals;-there is no point x at which both constraints bind simultaneously;-if gj(x) = 0 some j, then the derivative of g1is non-zero.

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