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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 $(4,2)$?

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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 $16.$

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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 $3x_1+2x_2$.

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