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What is the maximum value of $x^2+4xy-y^2$ for all $(x,y)$ satisfying $x^2+y^2 = 1$? [closed]

How do I determine the maximum value for a quadratic equation on an interval?Find the value of $p$ and $q$ of the quadratic equation.Given the equation of the parabola, for what values of $t$ will the quadratics have $0$, $1$, or $2$ solutions?How to deduce $,n^2+5n-12=0,Rightarrow, n^3 = 37n - 60$?Diophantine Equation or EllipseFor what values does $kx^2 - 6x - 4 = 0$ cut the x-axis?What is the value of $t$?What does changing the coefficients of polynomials graph?What are quadratic functions and what are they for?Finding the minimum value.

-1

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Does the trick have something to do with the equation of a circle?

quadratics

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edited Mar 21 at 4:56

Naman Kumar

22813

asked Mar 21 at 4:14

EdwardEdward

11

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closed as off-topic by Shailesh, Leucippus, jgon, Eevee Trainer, Saad Mar 21 at 5:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Leucippus, jgon, Eevee Trainer, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x=cos t,y=sin t$
  $endgroup$
  – lab bhattacharjee
  Mar 21 at 4:22
  

  
  
  
  

add a comment | 

-1

$begingroup$

Does the trick have something to do with the equation of a circle?

quadratics

share|cite|improve this question

edited Mar 21 at 4:56

Naman Kumar

22813

asked Mar 21 at 4:14

EdwardEdward

11

$endgroup$

closed as off-topic by Shailesh, Leucippus, jgon, Eevee Trainer, Saad Mar 21 at 5:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Leucippus, jgon, Eevee Trainer, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x=cos t,y=sin t$
  $endgroup$
  – lab bhattacharjee
  Mar 21 at 4:22
  

  
  
  
  

add a comment | 

$begingroup$

Does the trick have something to do with the equation of a circle?

quadratics

share|cite|improve this question

edited Mar 21 at 4:56

Naman Kumar

22813

asked Mar 21 at 4:14

EdwardEdward

11

$endgroup$

share|cite|improve this question

edited Mar 21 at 4:56

Naman Kumar

22813

asked Mar 21 at 4:14

EdwardEdward

11

share|cite|improve this question

edited Mar 21 at 4:56

Naman Kumar

22813

asked Mar 21 at 4:14

EdwardEdward

11

edited Mar 21 at 4:56

Naman Kumar

22813

edited Mar 21 at 4:56

Naman Kumar

22813

asked Mar 21 at 4:14

EdwardEdward

11

asked Mar 21 at 4:14

EdwardEdward

11

2 Answers
2

active

oldest

votes

1

$begingroup$

Hint. Let $x=sintheta, y=costheta$ to have $x^2+4xy-y^2=sin^2theta+4sinthetacostheta-cos^2theta$. Can you use calculus to minimise this function of one variable?

share|cite|improve this answer

answered Mar 21 at 4:21

YiFanYiFan

5,0161727

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why calculus $$2(sin2theta)-cos2thetalesqrt2^2+(-1)^2$$
  $endgroup$
  – lab bhattacharjee
  Mar 21 at 4:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That works as well. But if one didn't see that expression, calculus is the surefire way to go.
  $endgroup$
  – YiFan
  Mar 21 at 4:30
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $x^2+4xy-y^2=c$ and $dfrac c1=dfracx^2+4xy-y^2x^2+y^2$

$iff(c-1)(x/y)^2-4(x/y)+c+1=0$

As $x/y$ is real,the discriminant must be $ge0$

$$4^2ge4(c^2-1)iff c^2le5$$

The equality will occur if $dfrac xy=dfrac42(c-1)$

Alternatively,

Let $x^2+4xy-y^2=uimplies x^2-y^2=u-4xy$

$1^2=(x^2-y^2)^2+(2xy)^2=(u-4xy)^2+(2xy)^2iff20(xy)^2-8u(xy)+u^2-1=0$

The discriminant $(8u)^2-80(u^2-1)=16(5-u^2)$

share|cite|improve this answer

answered Mar 21 at 4:30

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

1

$begingroup$

Hint. Let $x=sintheta, y=costheta$ to have $x^2+4xy-y^2=sin^2theta+4sinthetacostheta-cos^2theta$. Can you use calculus to minimise this function of one variable?

share|cite|improve this answer

answered Mar 21 at 4:21

YiFanYiFan

5,0161727

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why calculus $$2(sin2theta)-cos2thetalesqrt2^2+(-1)^2$$
  $endgroup$
  – lab bhattacharjee
  Mar 21 at 4:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That works as well. But if one didn't see that expression, calculus is the surefire way to go.
  $endgroup$
  – YiFan
  Mar 21 at 4:30
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint. Let $x=sintheta, y=costheta$ to have $x^2+4xy-y^2=sin^2theta+4sinthetacostheta-cos^2theta$. Can you use calculus to minimise this function of one variable?

share|cite|improve this answer

answered Mar 21 at 4:21

YiFanYiFan

5,0161727

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why calculus $$2(sin2theta)-cos2thetalesqrt2^2+(-1)^2$$
  $endgroup$
  – lab bhattacharjee
  Mar 21 at 4:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That works as well. But if one didn't see that expression, calculus is the surefire way to go.
  $endgroup$
  – YiFan
  Mar 21 at 4:30
  

  
  
  
  

add a comment | 

$begingroup$

Hint. Let $x=sintheta, y=costheta$ to have $x^2+4xy-y^2=sin^2theta+4sinthetacostheta-cos^2theta$. Can you use calculus to minimise this function of one variable?

share|cite|improve this answer

answered Mar 21 at 4:21

YiFanYiFan

5,0161727

$endgroup$

share|cite|improve this answer

answered Mar 21 at 4:21

YiFanYiFan

5,0161727

answered Mar 21 at 4:21

YiFanYiFan

5,0161727

answered Mar 21 at 4:21

YiFanYiFan

5,0161727

0

$begingroup$

Let $x^2+4xy-y^2=c$ and $dfrac c1=dfracx^2+4xy-y^2x^2+y^2$

$iff(c-1)(x/y)^2-4(x/y)+c+1=0$

As $x/y$ is real,the discriminant must be $ge0$

$$4^2ge4(c^2-1)iff c^2le5$$

The equality will occur if $dfrac xy=dfrac42(c-1)$

Alternatively,

Let $x^2+4xy-y^2=uimplies x^2-y^2=u-4xy$

$1^2=(x^2-y^2)^2+(2xy)^2=(u-4xy)^2+(2xy)^2iff20(xy)^2-8u(xy)+u^2-1=0$

The discriminant $(8u)^2-80(u^2-1)=16(5-u^2)$

share|cite|improve this answer

answered Mar 21 at 4:30

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

0

$begingroup$

Let $x^2+4xy-y^2=c$ and $dfrac c1=dfracx^2+4xy-y^2x^2+y^2$

$iff(c-1)(x/y)^2-4(x/y)+c+1=0$

As $x/y$ is real,the discriminant must be $ge0$

$$4^2ge4(c^2-1)iff c^2le5$$

The equality will occur if $dfrac xy=dfrac42(c-1)$

Alternatively,

Let $x^2+4xy-y^2=uimplies x^2-y^2=u-4xy$

$1^2=(x^2-y^2)^2+(2xy)^2=(u-4xy)^2+(2xy)^2iff20(xy)^2-8u(xy)+u^2-1=0$

The discriminant $(8u)^2-80(u^2-1)=16(5-u^2)$

share|cite|improve this answer

answered Mar 21 at 4:30

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

$begingroup$

Let $x^2+4xy-y^2=c$ and $dfrac c1=dfracx^2+4xy-y^2x^2+y^2$

$iff(c-1)(x/y)^2-4(x/y)+c+1=0$

As $x/y$ is real,the discriminant must be $ge0$

$$4^2ge4(c^2-1)iff c^2le5$$

The equality will occur if $dfrac xy=dfrac42(c-1)$

Alternatively,

Let $x^2+4xy-y^2=uimplies x^2-y^2=u-4xy$

$1^2=(x^2-y^2)^2+(2xy)^2=(u-4xy)^2+(2xy)^2iff20(xy)^2-8u(xy)+u^2-1=0$

The discriminant $(8u)^2-80(u^2-1)=16(5-u^2)$

share|cite|improve this answer

answered Mar 21 at 4:30

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

share|cite|improve this answer

answered Mar 21 at 4:30

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 21 at 4:30

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 21 at 4:30

lab bhattacharjeelab bhattacharjee

228k15158279