For x ≥ 0 and y ≥ 0 satisfying x + y ≥ 6 and 2x + y ≥ 10, what is the smallest possible value of 4x + 3y?

The minimum of a linear expression over this unbounded region occurs at a corner. The constraints $x + y \geq 6$ and $2x + y \geq 10$ have corners at $(0, 10)$, $(6, 0)$, and the intersection of the two lines: subtracting $x + y = 6$ from $2x + y = 10$ gives $x = 4$, so $y = 2$, point $(4, 2)$. Now evaluate $4x + 3y$: at $(0,10)$ it is 30, at $(6,0)$ it is 24, and at $(4,2)$ it is $16 + 6 = 22$. The minimum is 22, choice C. Choice A (24) and choice B (30) are values at the other corners. Choice D (18) would need a point outside the region.