We can write the given problem as
$f(x_1, x_2, x_3 ) = - x_1^2 - x_2^2 – x_3^2 + 4x_1+6x_2 $
$h1(x_1, x_2 ) = x_1 +x_2 -2$
$h2(x_1, x_2 ) = 2x_1 + 3x_2 -12 $
Kuhn -Tucker conditions for maxima are
$ \frac{∂f}{(∂x_1 )}-λ_1 \frac{(∂h_1)}{(∂x_1 )}- λ_2 \frac{(∂h_2)}{(∂x_1 )} = 0, \,\,\,\ $
$∴ - 2x_1 +4 - λ_1 - 2 λ_2 = 0$ ……………………(1)
$ \frac{∂f}{(∂x_2 )}-λ_1 \frac{(∂h_1)}{(∂x_2 )}-λ_2 \frac{(∂h_2)}{(∂x_2 )} = 0, \,\,\,\ $
$∴ - 2x_2 + 6 - λ_1 - 3λ_2 = 0 $……………………(2)
$ \frac{∂f}{(∂x_3 )}-λ_1 \frac{ (∂h_1)}{(∂x_3 )}-λ_2 \frac{ (∂h_2)}{(∂x_3 )} = 0, \,\,\,\ $
$∴ x_3 = 0$ ………………....…………(3)
$ λ_1 h_1 =0, \,\,\,\ ∴ λ_1 (x_1 +x_2 -2)=0$ ………..(4)
$ λ_2 h_2 =0, \,\,\,\ ∴ λ_2(2x_1 + 3x_2 -12 ) =0$ ……………...(5)
$h_1(x_1, x_2) ≤0, \,\,\,\ ∴ x_1 +x_2 -2≤0$ ...……….(6)
$h_2(x_1, x_2) ≤0, \,\,\,\ ∴ 2x_1 + 3x_2 - 12≤0$ …….....(7)
$x_1, x_2, x_3 , λ_1 , λ_2≥ 0$ ………….(8)
Now depending upon the values of $λ_1$ & $λ_2$ following cases arise
Case1: $λ_1 =0$ and $λ_2=0$
In this case from (1) ,(2) and (3) ,we get
$ x_1 =2, x_2 =3, x_3 =0$
but these values , do not satisfy the conditions (6) &(7), Hence we reject
Case2: If $λ_1 =0$ and $λ_2≠0$
Since $λ_1$ =0 eqn(1) & (2) becomes $- 2x_1 +4 - 2 λ_2 = 0$ and $- 2x_2 + 6 - 3λ_2 = 0$ ,
eliminating $λ_2$ ,we get $2x_2 =3x_1$ ………………..(9)
Now from (5) $2x_1 + 3x_2 -12 =0$ ……………………(10)
(∵$λ_2≠0$)
Solving (9) & (10) , we get $x_1 =\frac{36}{13} , x_2 = \frac{3}{2} ,x_3=0$
But these values , also do not satisfy the conditions (6) ,(7) & (8)
Case3. If $λ_1≠ 0$ and $λ_2=0$
Since $λ_2 =0$ eqn(1) & (2) becomes $- 2x_1 +4 -λ_1 = 0$
and $-2x_2 + 6 -λ_1 = 0$ , eliminating $λ_1$ ,we get $-x_1 + x_2 =1$ ………………..(11)
Now from (4) $x_1 + x_2 = 2$ ……………………(12) (∵$λ_1≠0$)
Solving (9) & (10) , we get $x_1 = \frac{1}{2}, x_2 = \frac{3}{2}, x_3=0$ also from (1) $λ_1 =3$
Hence these values satisfy the conditions (6) , (7)
Therefore $x_1 =\frac{1}{2}, x_2 = \frac{3}{2} ,x_3=0$ is a feasible solution.
And $z_{Max} = \frac{17}{2}$
Case 4: If $λ_1 ≠0$ and $λ_2≠0$
If $λ_1 ≠0$ and $λ_2≠0$ , we get from (4) & (5)
$x_1 +x_2 -2=0$
$2x_1 + 3x_2 -12 =0$
Solving these equations ,we get,
$ x_1 = -6, x_2 = 8, x_3 = 0$
But these do not satisfy the non – negativity restrictions (8) .
Hence we reject the solution
∴ The solution is $x_1 = \frac{1}{2}, x_2 = \frac{3}{2} ,x_3=0$ & $z_{Max} = \frac{17}{2}$