Question

已知实数x.y满足3x＋4y－10＝0,求x²＋y²的最小值

Answer 1

3x+4y-10=0

x=(10-4y)/3

x^2+y^2=(10-4y)^2/9+y^2=(100-80y+16y^2)/9+y^2=(25y^2-80y+100)/9=(5y)^2-2*5*8y+64+36)/9

=[(5y+8)^2+36]/9

最小值是36/9=4