Maximize $z = 2x^2_1 +12x_1x_2 – 7x_2^2$

Subject to the constraints $2x_1 + 5x_2 ≤ 98$ and $x_1, x_2 ≥ 0.$

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2014

Let $ f(x_1, x_2) =2 x_1^2-7x_2^2+12(x_1- x_2)- \lambda(2x_1+5x_2-98)\\ =2 x_1^2-7x_2^2+12(x_1 x_2)- 2\lambda x_1+5\lambda x_2-98\lambda$

Kuhn-Tucker (k-4) Conditions are:

$ i) \dfrac {\partial L}{\partial x_1}=0\Rightarrow 4x_1+12x_2-2\lambda=0 \\ ii) \dfrac {\partial L}{\partial x_2}0\Rightarrow -14x_2+12x_1-5\lambda =0\\ iii) \lambda h(x_1,x_2)=0\Rightarrow \lambda(2x_1+5x_2-98)=0\\ iv) h(x_1x_2) \leq 0 \Rightarrow 2x_1+5x_2-98 \leq 0\\ v) \lambda,x_1,x_2 \geq 0$

From K-H condition (3) following two cases arises , Case 1: $\lambda =0$

From (1) $4x_1+12x_2=0$

From (2), $-14x_2+12x_1=0$

On solving simultaneously $x1=0, x2=0;$

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

LHS= $2(0) +5(0)-98=-98$

Values of $x_1$ & $x_2$ satisfies K-H condition (4) & (5)

$\therefore $ stationary point $X_0=(0,0)$

Optimal Solution at $X_0=Z_{max} \\ =2(0)^2-7(0)^2+12(0)(0) \\ =0$

Case 2:

From (3), $2x_1+5x_2-98=0 \\ \therefore 2x_1+5x_2=98------------------------- (6)$

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

$ x_1=44; x_2=2; \lambda=100$

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

$\therefore$ Stationary point $X_0=Z_{max} \\ =2(44)^2-7(2)^2+12(44)(2)=4900$

Hence, stationary point $X_0=(44,2)$ and $Z_{max}=4900.$