Question

求多项式x^2+4y^2-8x+12y+5的最小值

Answer 1

因为x^2+4y^2-8x+12y+5

=(x^2-8x+16)+(4y^2+12y+9)-20

=(x-4)^2+(2y+3)^2-20,

又因为(x-4)^2≥0,(2y+3)^2≥0,

所以x^2+4y^2-8x+12y+5=(x-4)^2+(2y+3)^2-20≥-20,

所以多项式x^2+4y^2-8x+12y+5的最小值-20.