Q110 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics

If $x$ and $y$ are two positive numbers such that $x+y=32$, then the minimum value of $x^2+y^2$ is,

  • A
    500
  • B
    256
  • C
    1024
  • Correct answer
    D
    512

Solution

$$ \begin{aligned} & \text { Let, } s=x^2+y^2 \\ & =x^2+(32-x)^2 \quad[\because x+y=32] \\ & \therefore \quad \frac{d s}{d x}=2 x+2(32-x)(-1) \\ & =2 x-2(32-x) \\ & =2 x-64+2 x \\ & =4 x-64 \\ & \end{aligned} $$ For maxima or minima, $$ \begin{array}{rlrl} \frac{d s}{d x} & =0 \\ \Rightarrow 4 x-64 & =0 \\ \Rightarrow \quad x & =16 \\ & \text { and } \quad y & =32-x=32-16=16 \end{array} $$ Again, $\frac{d^2 s}{d x}=4=+$ ive $\therefore s$ is minimum when $x=y=16$ $\therefore$ Minimum value of $$ s=16^2+16^2=256+256=512 $$

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

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