Question

Two numbers have a difference of 34. What is the sum of their squares if it is a minimum?

Answer 1

`a,b-the\ numbers\\\\a-b=34\to a=34+b\\\\a^2+b^2\to minimum\\\\\text{substitute:}\\\\(34+b)^2+b^2\to minimum\\\\f(b)=(34+b)^2+b^2\qquad\text{use}\ (x+y)^2=x^2+2xy+y^2\\\\f(b)=34^2+(2)(34)(b)+b^2+b^2\\\\f(b)=1156+68b+2b^2\to f(b)=2b^2+68b+1156\\\\y=ax^2+bx+c\\\\if\ a>0\ then\ a\ parabola\ op en\ up\\if\ a<0\ then\ a\ parabola\ op en\ down\\\\if\ a>0\ then\ a\ parabola\ has\ a\ minimum\ at\ a\ vertex\\if\ a<0\ then\ a\ parabola\ has\ a\ maximum\ at\ a\ vertex`

`\text{We have}\ a=2>0.\ \text{Therefore the parabola has the minimum at the vertex.}\\\\(h,\ k)-vertex\\\\h=(-b)/(2a);\ k=f(h)\\\\\text{We have}\ a=2\ \text{and}\ b=68.\ \text{Substitute:}\\\\h=(-68)/(2(2))=(-68)/(4)=-17\\\\k=f(-17)=2(-17)^2+68(-17)+1156=2(289)-1156+1156=578`

`\text{Therefore}\ b=-17\ \text{and}\ a=34+b\to a=34+(-17)=17.\\\\Answer:\ a^2+b^2=17^2+(-17)^2=289+289=578`