Solve: Minimize $z = 2x_1 + 3x_2$, such that
$x_1 + x_2 ≤ 4$
$3x_1 + x_2 ≥ 4$
$x_1 + 5x_2 ≥ 4$
and such that $0 ≤ x_1 ≤ 3$, and $0 ≤ x_2 ≤ 3.$
My attempt : From $x_1 + x_2 ≤ 4$ and $3x_1 + x_2 ≥ 4$, i got $x_1=0$ and $x_2=4$ .
From $x_1 + x_2 ≤ 4$ and $x_1 + 5x_2 ≥ 4$ i got $x_1=4, x_2=0$
After that im not able to proceed further
This is linear optimization with $2x_1+3x_2$ as objective. You have to first find the maximum value of $x_2$ possible within constraints(=feasible region) (because $x_2$ has the largest coefficient and our problem is maximization, also notice that the maximum of linear optimization is attained at corners of feasible region). Maximum value possible for $x_2$ is $x_2=3$ without violating any constraints. Then all we had to do next is finding maximum value of $x_1$ such that all constraints be satisfied. We see that $x_1 + x_2 ≤ 4$ is bottleneck. Thus $x_1=1$ is the best we could do and overall $\max(2x_1+3x_2)=11$.
In other words (if we like to think about it in "simplex method" terms) the maximum is attained at intersection $x_1+x_2=4, x_2=3$
According to linear programing a linear function $f(x_1,x_2)=2x_1+3x_2$ takes maximum or minimum value o\at the vertices of a convex polygon. Get the vertices ass pints of intersection of $x_1+x_2=4, 3x_1+x_2=4, x_1+5x_2=4, 1 \le x_1,x_2 \le 3.$ These conditions make a convex pentagon in $(x_1,x_2)$ plane whose vertices are $(3,1/3), (3, 1), (8/7,4/7), (1/3,3), (3,1/5)$ Then $f(x_1,x_2)=7,9,38/7,29/3,33/5,$ respectively, where $33/5$ is the required minimum value.
