written 6.3 years ago by | • modified 4.0 years ago |
Using Kuhn –Tucker condition to solve the following N.L.P.P.
Max. $Z = 10x_1 +4x_2 –2 x_1^2 –x_2^2, \text{subject to} \hspace{0.2cm} 2x_1 +x_2 ≤5$
$x_1, x_2 ≥ 0$
written 6.3 years ago by | • modified 4.0 years ago |
Using Kuhn –Tucker condition to solve the following N.L.P.P.
Max. $Z = 10x_1 +4x_2 –2 x_1^2 –x_2^2, \text{subject to} \hspace{0.2cm} 2x_1 +x_2 ≤5$
$x_1, x_2 ≥ 0$
written 6.2 years ago by | • modified 6.0 years ago |
We write the problem as
$f( x_1, x_2 ) = 10x_1 +4x_2 –2 x_1^2 –x_2^2$
and $h( x_1 , x_2) = 2x_1 +x_2 - 5 $
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
$10 – 4x_1 - 2λ = 0$ …….(1)
$4 – 2x_2 - λ = 0$……......(2)
$λ(2x_1 +x_2 - 5) = 0$...(3)
$2x_1 +x_2 - 5 ≤ 0$.…....(4)
$x_1 , x_2 , λ≥ 0$….............(5)
From (3) , we get either λ =0 or 2x$_1$ +x$_2$ - 5 = 0
Case (1):
If λ =0 ,then from (1) & (2) we get $x_1 = \frac{5}{2}$ & $x_2$ = 2 ,
but these values of $x_1$, $x_2$ does not satisfied eqn(4)
∴ λ = 0 does not give feasible solution
Case (2):
If $2x_1 + x_2 - 5 = 0$ …….(6)
Then from (1) $x_1 = \frac{(10 - 2λ)}{4}$ & from (2) $x_2 = \frac{(4 - λ)}{2}$,
putting both the values in (6) we get
$λ = \frac{4}{3}$, therefore $x_1 = \frac{11}{6}$ & $x_2 = \frac{4}{3}$
since these values satisfy all the conditions.
∴ Optimal sol. is $x_1 = \frac{11}{6}; x_2 = \frac{4}{3}$
∴ Max. z = $\frac{91}{6}$