2006 SMT/Algebra Problems/Problem 2

Problem

Find the minimum value of $2x^2+2y^2+5z^2-2xy-4yz-4x-2z+15$ for real numbers $x, y, z$.

Solution

Solution 1

Notice that $2x^2+2y^2+5z^2-2xy-4yz-4x-2z+15=(z-1)^2+(x-y)^2+(x-2)^2+(2z-y)^2+10$. This achieves a minimum value when all of the squares are $0$, that is, when $z-1=0, x-y=0, x-2=0,$ and $2z-y=0$. Solving, we find that $x=y=2$ and $z=1$ satisfy this, and so the minimum value is $\boxed{10}$.

Solution 2

Let $f(x,y,z)=2x^2+2y^2+5z^2-2xy-4yz-4x-2z+15$. The minimum value occurs when $\frac{\delta f}{\delta x}=\frac{\delta f}{\delta y}=\frac{\delta f}{\delta z}=0$. Taking these partial derivatives, we have

\[4x-2y-4=0\implies 2x-y=2\] \[4y-2x-4z=0\implies x-2y+2z=0\] \[10z-4y-2=0\implies 5z-2y=1\]

From the first and third equations, we have $x=\frac{y+2}{2}$ and $z=\frac{2y+1}{5}$. Plugging these into the second equation and solving, we find that $y=2$. From this we get $x=2$ and $z=1$. Therefore, the minimum value is $f(2,2,1)=\boxed{10}$.

See Also

2006 SMT/Algebra Problems