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

 Feb 3, 2019
 #2
avatar+104807 
+2

x^2 + y^2= 1      is a circle with a radius of 1 centered at the origin

 

Using implicit differentiation

 

(x + y)^2 =

 

x^2 + 2xy + y^2

 

2x  + 2y + 2xy' + 2yy'

 

y' [ 2x + 2y]  =  -2x - 2y

 

y'  =  -2 (x + y) / [ 2 (x + y ) ]

 

y' =  -1      

 

Ths is the slope of the tangent line to the circle at the points where the function is maxed

 

And this line will be perependicular to  the  line through the origin with a slope of 1  ( y = x )

 

But in a circle with a radius of 1 centered at the origin, this line will intersect the circle at  (√2/2, √2/2) and (-√2/2, -√2/ 2)

 

And using impiicit differentiation again on the circular function we get that

 

2x + 2yy' = 0

2yy' = - 2x

y' = -2x/2y  = -x/y

And this gives the slope of any  tangent line to the circle

Plugging in the given points will give us the slope of -1   which is what we want

 

So   (x + y)^2 is maxed at either (√2/2, √2/2) or (-√2/2, -√2/ 2)

 

And the max is    ( √2/2 +√2/2 )^2  =  ( √2 )^2  =   2

 

 

cool cool cool

 Feb 3, 2019
edited by CPhill  Feb 3, 2019
edited by CPhill  Feb 3, 2019
 #3
avatar+105606 
+1

Thanks Chris :)

Melody  Feb 3, 2019
 #4
avatar+104807 
0

OK....!!!

 

 

cool cool cool

CPhill  Feb 3, 2019
 #5
avatar+104807 
+1

Here's one more way

 

Let x =  rcos(theta)    and y = rsin(theta)

Since the circle ha a radius of 1......these become x = cos(theta)  and y = sin (theta)

 

So     let  A =  (x  + y)^2 =   x^2 + 2xy + y^2  =   cos^2(theta) + 2sin(theta)os(theta) + sin^2 (theta)  =

 

2sin(theta)cos(theta) + 1          take the derivative of this and set to 0

 

A' = 2sin^2(theta) - 2cos^2(theta) = 0

 

sin^2(theta) - cos^2(theta) = 0        factor

 

[sin (theta)  + cos(theta) ] [ sin (theta - cos(theta) ] = 0

 

Set both factors to 0 and solve for theta

 

sin (theta) + cos(theta) = 0                   sin (theta) - cos(theta) = 0

 

The solutions of the first = 3pi/4  and 7pi/4       these do not produce maxes

 

The solutions for the second =  pi/4 and 5pi/4

 

And x, y at these angles are   ( √2/2, √ 2/2)  and (-√2/2, -√2/2)

 

And these will produce the max for A  =    2

 

 

cool cool cool

 Feb 3, 2019
 #6
avatar+7725 
+1

Here is a way without calculus :)

 

Let x = sin θ, y = cos θ.

\(\displaystyle \max_{x,y\in\{x,y:x^2 +y^2 = 1\}} (x+y) = \max_{\theta \in [0,2\pi]} (\sin \theta + \cos \theta) = \max_{\theta \in [0,2\pi]}\left(\sqrt2 \sin\left(\theta+\dfrac{\pi}{4}\right)\right) = \sqrt2\).

 

So \(\displaystyle \max_{x,y\in\{x,y:x^2 +y^2 = 1\}} (x+y)^2 = \left(\sqrt2\right)^2 = 2\).

 Feb 4, 2019
 #7
avatar+23301 
+7

Maximum Function Help!
If \(x\) and \(y\) are real and \(x^2 + y^2 = 1\), compute the maximum value of \((x+y)^2\).

 

\(x^2 + y^2 = 1\)  is a circle with a radius of 1.

 

A circle can be defined as the locus of all points that satisfy the equations
\(x = r \cos(\alpha) \qquad y = r \sin(\alpha)\)
where x,y are the coordinates of any point on the circle, r is the radius of the circle and
\(\alpha\) is the parameter - the angle subtended by the point at the circle's center.

 

We substitute:

\(\begin{array}{|rcll|} \hline x^2 + y^2 &=& 1 \quad | \quad x = \cos(\alpha) \qquad y = \sin(\alpha) \\ \mathbf{\sin^2(\alpha) + \cos^2(\alpha)} & \mathbf{=}& \mathbf{ 1 } \\ \hline \end{array}\)

 

Maximul value:

\(\begin{array}{|rcll|} \hline (x+y)^2_\text{max} && \qquad | \quad x = \cos(\alpha) \qquad y = \sin(\alpha) \\ &=& \Big(\sin(\alpha) + \cos(\alpha)\Big)^2 \\ &=& \sin^2(\alpha) + \cos^2(\alpha) + 2\sin(\alpha)\cos(\alpha) \quad | \quad \sin^2(\alpha) + \cos^2(\alpha) = 1 \\ &=& 1+2\sin(\alpha)\cos(\alpha) \quad | \quad 2\sin(\alpha)\cos(\alpha) = \sin(2\alpha) \\ &=& 1+ \sin(2\alpha) \quad | \quad \text{The sin-function is max at 1 } \\ &=& 1+ 1 \\ (x+y)^2_\text{max}&\mathbf{=}& \mathbf{2} \\ \hline \end{array} \)

 

The maximum value of \((x+y)^2\) is 2

 

laugh

 Feb 7, 2019

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