Maths Olympiad Prep

Library / /24 of 75

Algebra Difficulty 4.6 AIME Prove it Italy

What is the minimum value of the expression x2−8xy+19y2−6y+14x^{2}-8 x y+19 y^{2}-6 y+14 as xx and yy vary over the real numbers?

Solution

The answer is 11. We can write
x2−8xy+19y2−6y+14==x2−8xy+16y2+3y2−6y+3+11==(x−4y)2+3(y−1)2+11≥11. \begin{aligned} & x^{2}-8 x y+19 y^{2}-6 y+14= \\ = & x^{2}-8 x y+16 y^{2}+3 y^{2}-6 y+3+11= \\ = & (x-4 y)^{2}+3(y-1)^{2}+11 \geq 11 . \end{aligned}
The value 11 is achieved for y=1,x=4y=1, x=4.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.