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Using Kuhn Tucker condition to the solve the following N.L.P.P.

Using Kuhn –Tucker condition to solve the following N.L.P.P.

Max. $Z = - x_1^2 - x_2^2 – x_3^2 + 4x_1+6x_2,$

$ \text{subject to} \hspace{0.2cm} x_1 + x_2 ≤2 ; 2x_1 + 3x_2 ≤12$

$x_1, x_2, x_3 ≥ 0$

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We can write the given problem as

$f(x_1, x_2, x_3 ) = - x_1^2 - x_2^2 – x_3^2 + 4x_1+6x_2 $

$h1(x_1, x_2 ) = x_1 +x_2 -2$

$h2(x_1, x_2 ) = 2x_1 + 3x_2 -12 $

Kuhn -Tucker conditions for maxima are

$ \frac{∂f}{(∂x_1 )}-λ_1 \frac{(∂h_1)}{(∂x_1 )}- λ_2 \frac{(∂h_2)}{(∂x_1 )} = 0, \,\,\,\ $

$∴ - 2x_1 +4 - λ_1 - 2 λ_2 = 0$ ……………………(1)

$ \frac{∂f}{(∂x_2 )}-λ_1 \frac{(∂h_1)}{(∂x_2 )}-λ_2 \frac{(∂h_2)}{(∂x_2 )} = 0, \,\,\,\ $

$∴ - 2x_2 + 6 - λ_1 - 3λ_2 = 0 $……………………(2)

$ \frac{∂f}{(∂x_3 )}-λ_1 \frac{ (∂h_1)}{(∂x_3 )}-λ_2 \frac{ (∂h_2)}{(∂x_3 )} = 0, \,\,\,\ $

$∴ x_3 = 0$ ………………....…………(3)

$ λ_1 h_1 =0, \,\,\,\ ∴ λ_1 (x_1 +x_2 -2)=0$ ………..(4)

$ λ_2 h_2 =0, \,\,\,\ ∴ λ_2(2x_1 + 3x_2 -12 ) =0$ ……………...(5)

$h_1(x_1, x_2) ≤0, \,\,\,\ ∴ x_1 +x_2 -2≤0$ ...……….(6)

$h_2(x_1, x_2) ≤0, \,\,\,\ ∴ 2x_1 + 3x_2 - 12≤0$ …….....(7)

$x_1, x_2, x_3 , λ_1 , λ_2≥ 0$ ………….(8)

Now depending upon the values of $λ_1$ & $λ_2$ following cases arise

Case1: $λ_1 =0$ and $λ_2=0$

In this case from (1) ,(2) and (3) ,we get

$ x_1 =2, x_2 =3, x_3 =0$

but these values , do not satisfy the conditions (6) &(7), Hence we reject

Case2: If $λ_1 =0$ and $λ_2≠0$

Since $λ_1$ =0 eqn(1) & (2) becomes $- 2x_1 +4 - 2 λ_2 = 0$ and $- 2x_2 + 6 - 3λ_2 = 0$ ,

eliminating $λ_2$ ,we get $2x_2 =3x_1$ ………………..(9)

Now from (5) $2x_1 + 3x_2 -12 =0$ ……………………(10)

(∵$λ_2≠0$)

Solving (9) & (10) , we get $x_1 =\frac{36}{13} , x_2 = \frac{3}{2} ,x_3=0$

But these values , also do not satisfy the conditions (6) ,(7) & (8)

Case3. If $λ_1≠ 0$ and $λ_2=0$

Since $λ_2 =0$ eqn(1) & (2) becomes $- 2x_1 +4 -λ_1 = 0$

and $-2x_2 + 6 -λ_1 = 0$ , eliminating $λ_1$ ,we get $-x_1 + x_2 =1$ ………………..(11)

Now from (4) $x_1 + x_2 = 2$ ……………………(12) (∵$λ_1≠0$)

Solving (9) & (10) , we get $x_1 = \frac{1}{2}, x_2 = \frac{3}{2}, x_3=0$ also from (1) $λ_1 =3$

Hence these values satisfy the conditions (6) , (7)

Therefore $x_1 =\frac{1}{2}, x_2 = \frac{3}{2} ,x_3=0$ is a feasible solution.

And $z_{Max} = \frac{17}{2}$

Case 4: If $λ_1 ≠0$ and $λ_2≠0$

If $λ_1 ≠0$ and $λ_2≠0$ , we get from (4) & (5)

$x_1 +x_2 -2=0$

$2x_1 + 3x_2 -12 =0$

Solving these equations ,we get, $ x_1 = -6, x_2 = 8, x_3 = 0$

But these do not satisfy the non – negativity restrictions (8) .

Hence we reject the solution

∴ The solution is $x_1 = \frac{1}{2}, x_2 = \frac{3}{2} ,x_3=0$ & $z_{Max} = \frac{17}{2}$

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