# Minimize $z=3x_1+2x_2$

Minimize $$z=3x_1+2x_2$$

s.t

$$2x_1+x_2\geq 10$$

$$-3x_1+2x_2\leq 6$$

$$x_1+x_2\geq 6$$

$$x_1,x_2 \geq 0$$

The feasible area is : link how can I find graphically that the answer is $$(4,2)$$?

Note that $$3x_1 + 2x_2 = (2x_1 + x_2) + (x_1 + x_2) \ge 10 + 6 = 16.$$
Solving the system of linear equations, equality holds at $$(4,2).$$ The other conditions are easily verified to be true for this point, so the minimum is $$16.$$
You want to find the minimal value of $$k$$ such that $$3x_1+2x_2=k$$ $$\therefore x_2=\frac{k}2-\frac32x_1$$ In order for a solution to exist the line $$x_2=\frac{k}2-\frac32x_1$$ must intersect the area shown. The value of $$k$$ can then be reduced which will translate the graph downwards until there is a single intersection with the given area. The value of $$k$$ when a single intersection occurs is then the minimal value of $$3x_1+2x_2$$.