Question

If $(x,y)$ lies on circle $x^2+y^2=1$, then the maximum values of $(x+y{)}^2$ is

• A. $3$
• B. $2$
• C. $3/2$
• D. $1/sqrt3$

Answer 1

The correct option is D $1/sqrt3$

We have

$(x,y)$ lies on the circle

$x^2+y^2=1......\left(1\right)$

Then,

Maximum values of ${(x+y)}^2$ is

${(x+y)}^2=x^2+y^2+2xy$

$=1+2xy$

$\text{let}$

$z=1+2xy$

By equation (1) to

$y=\sqrt{1-x^2}$

$z=1+2x\sqrt{1-x^2}$

On diff. and we get.

$\frac{dz}{dx}=0+2[x\frac{d}{dx}\sqrt{1-x^2}+\sqrt{1-x^2}]$

$\frac{dz}{dx}=2x\times \frac{1}{2\sqrt{1-x^2}}(-2x)+\sqrt{1-x^2}$

For maximum and minimum

$\frac{dz}{dx}=0$

$\frac{-4x^2}{2\sqrt{1-x^2}}+\sqrt{1+x^2}=0$

$\Rightarrow \frac{-2x^2}{\sqrt{1-x^2}}+\sqrt{1+x^2}=0$

$\Rightarrow -2x^2+(1-x^2)=0$

$\Rightarrow -2x^2+1-x^2=0$

$\Rightarrow -3x^2+1=0$

$\Rightarrow -3x^2=-1$

$\Rightarrow x^2=\frac{1}{3}$

$\Rightarrow x=\frac{1}{\sqrt{3}}$