Q109 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Mathematics

If $x+y=k, x>0, y>0$, then $x^2+y^2$ is minimum, if

  • A
    $x>y$
  • B
    $x < y$
  • Correct answer
    C
    $x=y$
  • D
    $x=2 y$

Solution

$$ \begin{gathered} x+y=k \\ \Rightarrow y=k-x, x>0, y>0 \end{gathered} $$ Consider $$ \begin{aligned} & A=x^2+y^2=x^2+(k-x)^2=2 x^2-2 k x+k^2 \\ & \frac{d A}{d x}=4 x-2 k=0 \Rightarrow x=\frac{k}{2} \\ & \frac{d^2 A}{d x^2}=4>0 \end{aligned} $$ $\therefore$ At $x=y=\frac{k}{2}$, we have Minimum vale of $x^2+y^2$.

Hence, the correct answer is option: 'C'.

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