2017/18 AHSMC Part 1 Problem 9

Let $A$, $B$ be real positive numbers. Then $$A(3x + y + 2z) + B(2y - x + 4z) \ge 3A + 5B \iff (3A - B)x + (A + 2B)y + (2A + 4B)z \ge 3A + 5B.$$ If take $3A - B = 7$, $A + 2B = 5$, $2A + 4B = 10$, which is equivalent to $A = \dfrac{19}{7}$, $B = \dfrac{8}{7}$, the above inequality can be written as $$7x + 5y + 10z \ge \dfrac{97}{7}.$$ Hence the minimum value of $7x + 5y + 10z$ is $\dfrac{97}{7}$. We may take $x = \dfrac{1}{7}$, $y = \dfrac{18}{7}$, $z = 0$ to justify that the minimum value is attainable. The answer is (B).