written 3.8 years ago by | modified 18 months ago by |

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

Max. $Z = 10x_1 +4x_2 –2 x_1^2 –x_2^2, \text{subject to} \hspace{0.2cm} 2x_1 +x_2 ≤5$

$x_1, x_2 ≥ 0$

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

written 3.8 years ago by | modified 18 months ago by |

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

Max. $Z = 10x_1 +4x_2 –2 x_1^2 –x_2^2, \text{subject to} \hspace{0.2cm} 2x_1 +x_2 ≤5$

$x_1, x_2 ≥ 0$

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written 3.7 years ago by | modified 3.5 years ago by |

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 = $\frac{91}{6}$

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