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 $f(2,2,1)=\boxed{10}$ .

Page

Toolbox

Search

2006 SMT/Algebra Problems/Problem 2

Contents

Problem

Solution

Solution 1

Solution 2

See Also