Question: Chapter 6 .
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Using Kuhn-Tucker condition, solve the following NLPP.

Maximise, $z=2x_{1}^{2}-7x_{2}^{2}+12x_{1}x_{2}$

Subject to, $2x_{1}+5x_{2} \le 98$

$x_{1}, \ x_{2} \ge 0$

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modified 6 weeks ago  • written 6 weeks ago by gravatar for tanya.tanyabarnwal tanya.tanyabarnwal0
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Solution:

$z=2x_{1}^{2}-7x_{2}^{2}+12x_{1}x_{2}$

Subject to, $2x_{1}+5x_{2} \le 98$

$x_{1}, \ x_{2} \ge 0$

We write the given problem as

$f(x_{1}, \ x_{2})=2x_{1}^{2}-7x_{2}^{2}+12x_{1}x_{2}$ -------------(1)

$h(x_{1}, \ x_{2})=2x_{1}+5x_{2}-98$ ----------------(2)

Now, Kuhn-Tucker condition are,

$\cfrac{\delta f}{\delta x_{1}}-\lambda \cfrac{\delta h}{\delta x_{1}}= 0$ -------------(3)

$\cfrac{\delta f}{\delta x_{2}}-\lambda \cfrac{\delta h}{\delta x_{2}}= 0$ -------------(4)

$\lambda h(x_{1}, \ x_{2})=0, \ h(x_{1}, \ x_{2}) \le 0, \ \lambda \ge 0$ -------------(5)

Solve equation (1) and (2) using equation (3), we get,

$(4x_{1}-0+12x_{2})-\lambda (2+0-0)=0 \Longrightarrow 4x_{1}+12x_{2}-2 \lambda=0$ ----------------(6)

Solve equation (1) and (2) using equation (4), we get,

$(0-14x_{2}+12x_{1})-\lambda (0+5-0)=0 \Longrightarrow -14x_{2}+12x_{1}-5 \lambda=0$ ----------------(7)

Using equation (5) we get,

$\lambda (2x_{1}+5x_{2}-98)=0$ ---------------(8)

$2x_{1}+5x_{2}-98 \le 0$ -------------------(9)

$x_{1}, \ x_{2}, \ \lambda \ge 0$ ------------------(10)

From equation (8), we get,

$\lambda =0 \ or \ (2x_{1}+5x_{2}-98)=0$

Case 1: If $\lambda = 0 $

From (6) and (7) we get,

$4x_{1}+12x_{2} =0 $

$-14x_{2}+12x_{1}=0$

By solving the equation simulatneously we get, $x_{1}=0, \ x_{2}=0$

This solution gives $z=0$.

Hence, for $\lambda=0$, feasible solution is not obtained.

$\therefore$ We reject these values.

Case 2: $2x_{1}+5x_{2}-98=0$

To find $x_{1}, \ x_{2}$ we obtain one more relation between $x_{1}, \ x_{2}$ by eliminating $\lambda$ from equation (6) and equation (7),

$4x_{1}+12x_{2}=0$ --------------(11)

$-14x_{2}+12x_{1}=0$ ---------------(12)

Now multiply equation (11) by 5 and multiply equation (12) by 2 and subtract,

$20x_{1}+60x_{2}=0$

$28x_{2}+24x_{1}=0$

Subtract both the equation,

$20x_{1}+60x_{2}=0 \\ 24x_{1}-28x_{2}=0 \\ - \ \ \ \ \ + \\ -------- \\ -4x_{1}+88x_{2}=0$

$-x_{1}+22x_{2}=0$

$x_{1}=22x_{2}$ ------------(13)

Put $x_{1}=22x_{2}$ in equation $2x_{1}+5x_{2}-98=0$

$\therefore 2(22x_{2})+5x_{2}-98=0$

$44x_{2}+5x_{2}=98$

$49x_{2}=98$

$x_{2}=2$

Put $x_{2}=2$ in (13),

$x_{1}=22(2)=44$

Now from equation (6),

$4(44)+12(2)-2 \lambda=0$

$176+24-2\lambda=0$

$200=2\lambda$

$\lambda=100, \ \lambda \gt 0$

These value satisfy all the necessary condition.

$\therefore$ The optimal solution is $x_{1}=44, \ x_{2}=2$

Put $x_{1} \ \& \ x_{2}$ value in equation (1),

$\therefore z_{max}=2(1936)-7(4)+12(44)(2)=4900$

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