Maximize $z = 8x_1 + 10x_2 – x_1^2 – x_2^2$

Subject to $3x_1 + 2x_2 ≤ 6 \\ X_1, x_2 ≥ 0 $

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : MAY 2014

Let $f(x_1,x_2)=8x_1+10x_2-x_1^2-x_2^2;$

$h=3x_1+2x_2-6$ and $\lambda$ be Lagrangian Multiplier

$\therefore $ Lagrangian function $L=f- \lambda h\\ \therefore L=(8x_1+10x_2-x_1^2-x_2^2)-\lambda(3x_1+2x_2-6)\\ =8x_1+10x_2-x_1^2+x_2^2-3\lambda x_1 -2\lambda x_2+6 \lambda$

Kuhn_Tucker (K-H) conditions are,

$i)\dfrac {\partial L}{\partial x_2}= 0\Rightarrow 10-2x_2-2\lambda =0\\ \therefore 0x_1+2x_2+2\lambda =10 \\ ii) \dfrac {\partial L}{\partial x_1}= 0\Rightarrow 8-2x_1-3\lambda =0\\ \therefore 2x_1+0x_2+3\lambda =8 \\ iii) \lambda h (x_1,x_2)=0\Rightarrow \lambda (3x_1+2x_2-6)=0\\ iv) h(x_1,x_2)=0\Rightarrow (3x_1+2x_2-6) \leq 0 \\ v) \lambda , x_1,x_2\geq 0$

from K-H condition (3), following 2 cases arises

Case 1:

$\lambda =0$

From (1) , $2x_1=8 \therefore x_1=4$

From (2), $2x_2=10 \therefore x_2=5 $

Substitute $x_1$ and $x_2$ in (4)

$$LHS=3(4)+2(5)-6=16 ≠0$$

Value of $x_1$ and $x_2$ do not satisfy K-H condition (4)

$\therefore $ we reject this case.

Case 2:$\lambda ≠0$

From (3), $3x_1+2x_2-6=0 \\ 3x1+2x2+0=0\lambda =6------------------------ (6)$

Solving (1),(2) and (6) Simultaneously we get

$x_1=\dfrac 4{13};x_2=\dfrac {33}{13}; \lambda =\dfrac {32}{13} $

values of $x_1$ and $x_2$ satisfies K-H condition (4) and (5)

$\therefore $ Stationary point $x_0=\Bigg(\dfrac 4{13},\dfrac {33}{13}$

since $\lambda \gt 0$, Z is maximum at $X_0$

Maximum value of $X_0=Z_{max}$

$$=8\times \dfrac 4{13}+10 \times \dfrac {33}{13}-\Bigg(\dfrac 4{13}\Bigg)^2 -\Bigg(\dfrac {32}{13}\Bigg)^2 \\ =\dfrac {277}{13} \\ =21.3077$$