, 2014

Bottle A contains 40 g of which 10% is acid.

Thus, it contains $0.1 \times 40 = 4$ g of acid and $40-4 = 36$ g of water.

Bottle B contains 50 g of which 20% is acid.

Thus, it contains $0.2 \times 50 = 10$ g of acid and $50-10 = 40$ g of water.

Bottle C contains 50 g of which 30% is acid.

Thus, it contains $0.3 \times 50 = 15$ g of acid and $50-15 = 35$ g of water.

In total, the three bottles contain $40+50+50=140$ g, of which $4+10+15=29$ g is acid and $140-29=111$ g is water.

The new mixture has mass 60 g of which 25% is acid.

Thus, it contains $0.25 \times 60 = 15$ g of acid and $60 - 15 = 45$ g of water.

Since the total mass in the three bottles is initially 140 g and the new mixture has mass 60 g, then the remaining contents have mass $140 - 60 = 80$ g.

Since the total mass of acid in the three bottles is initially 29 g and the acid in the new mixture has mass 15 g, then the acid in the remaining contents has mass $29 - 15 = 14$ g.

This remaining mixture is thus $\text{14 g 80 g 100 = 17.5}$ acid.
Since $3x+4y=10$, then $4y = 10 - 3x$.

Therefore, when $3x+4y=10$, $$\begin{aligned} x^2 + 16y^2 &= x^2 + (4y)^2 \\ &= x^2 + (10-3x)^2\\ &= x^2 + (9x^2 - 60x + 100)\\ &= 10x^2 - 60x+100\\ &= 10(x^2 - 6x + 10)\\ &= 10(x^2-6x+9+1)\\ &= 10((x-3)^2+1)\\ &= 10(x-3)^2 + 10\end{aligned}$$ Since $(x-3)^2 \geq 0$, then the minimum possible value of $10(x-3)^2 + 10$ is $10(0)+10 = 10$. This occurs when $(x-3)^2 = 0$ or $x=3$.

Therefore, the minimum possible value of $x^2+16y^2$ when $3x+4y=10$ is 10.