Maximize $z = 10x_1 + 4x_2 – 2x_2^2 – x_2^2$

Subject to $2x_1 + x_2 ≤ 5$ and $x_1, x_2 ≥ 0 $

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2015

The problem in standard way

$F(x_1, x_2) = 10x_1 + 4x_2 – 2x_1^2 –x_2^2$

And $h (x_1, x_2) = 0, h (x_1, x_2) \lt 0$

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) vwe get either $λ = 0$ or $2x_1 + x_2 – 5 = 0 $

Case1: If $λ = 0$ from (1) and (2)

We get $4x_1 = 10, x_1 = 5/2$

And $2 \times 2 = 4 \space \space i.e \space \space xz = 2$

Putting these values of $x_1$ and $x_2$ in LHS of (4) we get

LHS $= 5 + 2 – 5 = 2 ≤ 0$

Thus these values ndo not satisfy (4) Hence $λ = 0$ does not yield a feasible solution we reject these values

Case 2: if $λ ≠ 0 \space 2x_1 + x_2 – 5 = 0 $

We no solve (1), (2), and (6)

Subtracting twice the second equation from the first.

$10 – 4x_1 – 8 + 4x_2 = 0$ so $2x_1 – 2x_2 =1 $

Multiply (6) by 2 $4x_1 +2x_2 = 10$

By adding we get $6x_1 = 11$ so $x_1 = 11/6$

$$X_2 = 5 – 2x_1 = 4/3$$

From (2) $x = 4 – 2x_2 = 4 – 8/3 = 4/3$

Thus satisfies all the necessary condition

The optimal solution is $x_1 = 11/6 \space \space x_2 = 4/3$

$Z_{max} = 10 \times 11/6 + 4 \times 4/3 – 2 (11/6)^2 – (4/3)^2 = 91/6 $