## zmudz one year ago If $$x$$ and $$y$$ are real and $$x^2 + y^2 = 1$$, compute the maximum value of $$(x+y)^2.$$

1. triciaal

|dw:1441474466693:dw|

2. imqwerty

hint-this is the equation of a circle with center at origin

3. IrishBoy123

lagrange multiplier

4. anonymous

$x^2+y^2=1 \Rightarrow y^2 = 1 - x^2$ $(x+y)^2 = x^2+2xy+y^2=x^2+y^2+2xy=1+2xy=1+2x \sqrt{1-x^2}$to maximize, differentiate w.r.t. x =>$\frac{ d }{ dx }\left[ \left( x+y \right)^2 \right]=\frac{ d }{ dx }\left[ 1+2x\sqrt{1-x^2} \right]=2\sqrt{1-x^2}-2x\cdot \frac{ 2x }{ \sqrt{1-x^2} }=0$=>$2(1-x^2)-4x = 0 \Rightarrow x^2+2x-1=0 \Rightarrow x= -1\pm \sqrt2$

5. ganeshie8

or you may use AM-GM inequality \begin{align}(x+y)^2 &= x^2 + y^2 + \color{red}{2xy}\\~\\ &\le x^2 + y^2 + \color{red}{x^2+y^2}\\~\\ &= 1+1 \end{align}

6. anonymous

AM-GM?

7. imqwerty

(x-0)^2 +(y-0)^2 = sqrt(1) center=(0,0) radius r = 1 so u can use parametric form to denote any point x,y which lies on it like- x=x1 + rcos(theta) y=y1+rsin(theta) |dw:1441474744572:dw| x1,y1=0,0 so x= rcos(theta) y=rsin(theta) nd u need to find max of (x+y)^2 $(rcos \theta+rsin \theta)^2$ rcos(theta) + rsin(theta) ranges frm -sqrt(r) to + sqrt(r) so max value of rcostheta +rsintheta gives max of (x+y)^2 nd we knw r=1 so (1)^2 =1

8. ganeshie8

fir $$x,y\in \mathbb{R}$$ we have : $$(x-y)^2 \ge 0 \\\implies x^2+y^2 -2xy \ge 0 \implies 2xy \le x^2+y^2$$ so, \begin{align}(x+y)^2 &= x^2 + y^2 + \color{red}{2xy}\\~\\ &\le x^2 + y^2 + \color{red}{x^2+y^2}\\~\\ &= 1+1 \end{align}

9. anonymous

if x = sqrt(2)/2 = y then (x+y)^2 = 2

10. anonymous

1 is not max.

11. anonymous

2 is max.

12. anonymous

what does AM-GM stand for?

13. imqwerty

14. ganeshie8
15. anonymous

max of rcostheta + rsintheta = sqrt(2)/2 + sqrt(2)/2 = sqrt(2) => max = (sqrt(2))^2 = 2

16. imqwerty

aww not again!! thanks :)

17. imqwerty

so max of of rcos(theta)+rsin(theta) will not be 1 :P it is root(2) nd pluggin it in (x+y)^2 u get 2 as the answer :)

18. IrishBoy123

$$g(x,y) = x^2 + y^2 - 1$$ $$f(x,y) = (x+y)^2 = 1 +2xy$$ $$\nabla f = <2y, 2x>$$ $$\nabla g = <2x, 2y>$$ $$\nabla f = \lambda \nabla g \implies \lambda = \frac{y}{x} = \frac{x}{y}$$ $$\implies x^2 = y^2$$ $$2x^2 = 1, \ \ \ \ x = \frac{1}{\sqrt{2}} = y$$ $$f_{max} = 2$$