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

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

Max. $Z = 8x_1 +10x_2 – x_1^2 –x_2^2, \text{subject to} \hspace{0.2cm} 3x_1 +2x_2 ≤6$

$x_1, x_2 ≥ 0$

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We write the problem as

$f( x_1, x_2 ) = 8x_1 +10x_2 – x_1^2 – x_2^2$

and $h( x_1 , x_2) = 3x_1 +2x_2 - 6$

Now , Kuhn –Tucker conditions are

$\frac{∂f}{(∂x_1 )}-λ \frac{∂h}{(∂x_1 )} = 0, \frac{∂f}{(∂x_2 )}-λ \frac{∂h}{(∂x_2 )} = 0 , h(x_1, x_2) ≤0$

$λh(x_1, x_2) =0 , λ≥0$

∴ we get

$8 – 2x_1 - 3λ = 0$ …………………(1)

$10 – 2x_2 - 2λ = 0$ ……………….(2)

$λ(3x1 +2x2 - 6) =0$………….(3)

$3x_1 +2x_2 -6≤0$ ………………(4)

$x_1 , x_2 , λ≥ 0$………………………..(5)

From (3) , we get either $λ =0$ or $3x_1 +2x_2 - 6 =0$

Case (1): If $λ =0$ ,then from (1) & (2) we get $x_1 =4$ & $x_2 =5$ ,but these values of $x_1, x_2$ does not satisfied eqn(4)

∴ λ =0 does not gives feasible solution

Case (2) If $3x_1 +2x_2 - 6 =0$ ……………………………….(6)

Then from (1), $x_1 = \frac{(8 - 3 λ)}{2}$ & from (2), $x_2 = \frac{(10 - 2 λ)}{2}$ .Putting both the values in (6) we get
$λ = \frac{32}{13}$. Therefore $x_1 = \frac{4}{13}$ & $x_2 = \frac{33}{13}$, since these values satisfy all the conditions.

∴ Optimal sol. is $x_1 = \frac{4}{13}$ & $x_2 = \frac{33}{13}$

∴ $Max. Z = \frac{277}{13}$