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

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triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441474466693:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2hintthis is the equation of a circle with center at origin

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1lagrange multiplier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+y^2=1 \Rightarrow y^2 = 1  x^2\] \[(x+y)^2 = x^2+2xy+y^2=x^2+y^2+2xy=1+2xy=1+2x \sqrt{1x^2}\]to maximize, differentiate w.r.t. x =>\[\frac{ d }{ dx }\left[ \left( x+y \right)^2 \right]=\frac{ d }{ dx }\left[ 1+2x\sqrt{1x^2} \right]=2\sqrt{1x^2}2x\cdot \frac{ 2x }{ \sqrt{1x^2} }=0\]=>\[2(1x^2)4x = 0 \Rightarrow x^2+2x1=0 \Rightarrow x= 1\pm \sqrt2 \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1or you may use AMGM inequality \[\begin{align}(x+y)^2 &= x^2 + y^2 + \color{red}{2xy}\\~\\ &\le x^2 + y^2 + \color{red}{x^2+y^2}\\~\\ &= 1+1 \end{align}\]

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2(x0)^2 +(y0)^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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1fir \(x,y\in \mathbb{R}\) we have : \((xy)^2 \ge 0 \\\implies x^2+y^2 2xy \ge 0 \implies 2xy \le x^2+y^2 \) so, \[\begin{align}(x+y)^2 &= x^2 + y^2 + \color{red}{2xy}\\~\\ &\le x^2 + y^2 + \color{red}{x^2+y^2}\\~\\ &= 1+1 \end{align}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if x = sqrt(2)/2 = y then (x+y)^2 = 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what does AMGM stand for?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2what mistake did i made?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0max of rcostheta + rsintheta = sqrt(2)/2 + sqrt(2)/2 = sqrt(2) => max = (sqrt(2))^2 = 2

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2aww not again!! thanks :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2so 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 :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(g(x,y) = x^2 + y^2  1\) \(f(x,y) = (x+y)^2 = 1 +2xy\) \(\nabla f = <2y, 2x>\) \(\nabla g = <2x, 2y>\) \(\nabla f = \lambda \nabla g \implies \lambda = \frac{y}{x} = \frac{x}{y}\) \(\implies x^2 = y^2\) \(2x^2 = 1, \ \ \ \ x = \frac{1}{\sqrt{2}} = y\) \(f_{max} = 2\)
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