Global extrema of $f(x,y)=3x^2-y^3$ on the circle $x^2+y^2 leq 25$












0












$begingroup$


I wanted to know if my method and my results are correct. There isn't a solution to that question, so I don't know if it's correct.



Given $f(x,y)=3x^2-y^3$ and I need to find its global extremas on the circle $x^2+y^2 leq 25$.



So inside the domain, I have $nabla f =(6x, -3y^2)= (0,0)$ which yields the trivial case $x=y=0$
Now on the border, we have $x^2+y^2=25$. I use lagrange multiplier : $nabla f = lambda nabla g$ where $g(x,y) =x^2+y^2-25$

This gives $(6x, -3y^2)= lambda (2x, 2y)$.

If $x neq 0$ and $y neq 0$, then $lambda = 3$, $y=-2$ and $x= pm sqrt{21}$. If $x=0, y = pm 5$. If $y=0, x = pm 5$. So if we plug our values into our function, we get $0, 71, -125, 125, 75$ So there is a global maximum of 125 at $(0,-5)$ and a global minimum of $-125$ at $(0, 5)$.
Are my solutions correct ?

Also, I guess I could have used polar coordinates on the border given that it's a circle. But if we do that ($r=sqrt{25}=5$) and try to take the derivative of our function in polar coordinates, we get $5^2frac{d}{d theta}(3cos^2(theta)-5sin^3(theta))$ which isn't nice to derivate. Do you agree that it's much simpler to use Lagrange multiplier in this case ?



Thanks for your help !






real-analysis calculus analysis multivariable-calculus lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    Typo: ... the circle $x^2 + y^2 le 25$.
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 12:44












  • $begingroup$
    @PaulFrost Uh, typo sorry. Thanks for pointing it out !
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:44


















0












$begingroup$


I wanted to know if my method and my results are correct. There isn't a solution to that question, so I don't know if it's correct.



Given $f(x,y)=3x^2-y^3$ and I need to find its global extremas on the circle $x^2+y^2 leq 25$.



So inside the domain, I have $nabla f =(6x, -3y^2)= (0,0)$ which yields the trivial case $x=y=0$
Now on the border, we have $x^2+y^2=25$. I use lagrange multiplier : $nabla f = lambda nabla g$ where $g(x,y) =x^2+y^2-25$

This gives $(6x, -3y^2)= lambda (2x, 2y)$.

If $x neq 0$ and $y neq 0$, then $lambda = 3$, $y=-2$ and $x= pm sqrt{21}$. If $x=0, y = pm 5$. If $y=0, x = pm 5$. So if we plug our values into our function, we get $0, 71, -125, 125, 75$ So there is a global maximum of 125 at $(0,-5)$ and a global minimum of $-125$ at $(0, 5)$.
Are my solutions correct ?

Also, I guess I could have used polar coordinates on the border given that it's a circle. But if we do that ($r=sqrt{25}=5$) and try to take the derivative of our function in polar coordinates, we get $5^2frac{d}{d theta}(3cos^2(theta)-5sin^3(theta))$ which isn't nice to derivate. Do you agree that it's much simpler to use Lagrange multiplier in this case ?



Thanks for your help !






real-analysis calculus analysis multivariable-calculus lagrange-multiplier







share|cite|improve this question











$endgroup$












  • $begingroup$
    Typo: ... the circle $x^2 + y^2 le 25$.
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 12:44












  • $begingroup$
    @PaulFrost Uh, typo sorry. Thanks for pointing it out !
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:44
















0












0








0





$begingroup$


I wanted to know if my method and my results are correct. There isn't a solution to that question, so I don't know if it's correct.



Given $f(x,y)=3x^2-y^3$ and I need to find its global extremas on the circle $x^2+y^2 leq 25$.



So inside the domain, I have $nabla f =(6x, -3y^2)= (0,0)$ which yields the trivial case $x=y=0$
Now on the border, we have $x^2+y^2=25$. I use lagrange multiplier : $nabla f = lambda nabla g$ where $g(x,y) =x^2+y^2-25$

This gives $(6x, -3y^2)= lambda (2x, 2y)$.

If $x neq 0$ and $y neq 0$, then $lambda = 3$, $y=-2$ and $x= pm sqrt{21}$. If $x=0, y = pm 5$. If $y=0, x = pm 5$. So if we plug our values into our function, we get $0, 71, -125, 125, 75$ So there is a global maximum of 125 at $(0,-5)$ and a global minimum of $-125$ at $(0, 5)$.
Are my solutions correct ?

Also, I guess I could have used polar coordinates on the border given that it's a circle. But if we do that ($r=sqrt{25}=5$) and try to take the derivative of our function in polar coordinates, we get $5^2frac{d}{d theta}(3cos^2(theta)-5sin^3(theta))$ which isn't nice to derivate. Do you agree that it's much simpler to use Lagrange multiplier in this case ?



Thanks for your help !






real-analysis calculus analysis multivariable-calculus lagrange-multiplier







share|cite|improve this question











$endgroup$




I wanted to know if my method and my results are correct. There isn't a solution to that question, so I don't know if it's correct.



Given $f(x,y)=3x^2-y^3$ and I need to find its global extremas on the circle $x^2+y^2 leq 25$.



So inside the domain, I have $nabla f =(6x, -3y^2)= (0,0)$ which yields the trivial case $x=y=0$
Now on the border, we have $x^2+y^2=25$. I use lagrange multiplier : $nabla f = lambda nabla g$ where $g(x,y) =x^2+y^2-25$

This gives $(6x, -3y^2)= lambda (2x, 2y)$.

If $x neq 0$ and $y neq 0$, then $lambda = 3$, $y=-2$ and $x= pm sqrt{21}$. If $x=0, y = pm 5$. If $y=0, x = pm 5$. So if we plug our values into our function, we get $0, 71, -125, 125, 75$ So there is a global maximum of 125 at $(0,-5)$ and a global minimum of $-125$ at $(0, 5)$.
Are my solutions correct ?

Also, I guess I could have used polar coordinates on the border given that it's a circle. But if we do that ($r=sqrt{25}=5$) and try to take the derivative of our function in polar coordinates, we get $5^2frac{d}{d theta}(3cos^2(theta)-5sin^3(theta))$ which isn't nice to derivate. Do you agree that it's much simpler to use Lagrange multiplier in this case ?



Thanks for your help !






real-analysis calculus analysis multivariable-calculus lagrange-multiplier




real-analysis calculus analysis multivariable-calculus lagrange-multiplier






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 7 at 19:20







Poujh

















asked Dec 19 '18 at 12:41









PoujhPoujh

569516




569516












  • $begingroup$
    Typo: ... the circle $x^2 + y^2 le 25$.
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 12:44












  • $begingroup$
    @PaulFrost Uh, typo sorry. Thanks for pointing it out !
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:44




















  • $begingroup$
    Typo: ... the circle $x^2 + y^2 le 25$.
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 12:44












  • $begingroup$
    @PaulFrost Uh, typo sorry. Thanks for pointing it out !
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:44


















$begingroup$
Typo: ... the circle $x^2 + y^2 le 25$.
$endgroup$
– Paul Frost
Dec 19 '18 at 12:44






$begingroup$
Typo: ... the circle $x^2 + y^2 le 25$.
$endgroup$
– Paul Frost
Dec 19 '18 at 12:44














$begingroup$
@PaulFrost Uh, typo sorry. Thanks for pointing it out !
$endgroup$
– Poujh
Dec 19 '18 at 12:44






$begingroup$
@PaulFrost Uh, typo sorry. Thanks for pointing it out !
$endgroup$
– Poujh
Dec 19 '18 at 12:44












1 Answer
1






active

oldest

votes


















1












$begingroup$

Your solutions are correct and Lagrange is simpler than polar coordinates.



Another simple method: from $g(x,y)=0$ we get $x^2=25-y^2$, hence $f(x,y)=75-3y^2-y^3$.



Now discuss the function $h(y)=75-3y^2-y^3$ on the intervall $[-5,5]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So I can take the derivative of $75-3y^2-y^3$, set it equal to $0$ and solve for $y$ appropriately. Then solve for $x=pm sqrt{25-y^2}$. I get $-3y(2+y)=0$ which yields again $y=-2$ implies $x= pm sqrt{21}$ and $y=0$ implies $x=pm 5$. But in this case, how do I get the case $x=0, y = pm 5$ ?
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:58






  • 1




    $begingroup$
    The cases $x=0, y= pm 5$ are not obtained with the derivative, since $h'( pm 5) ne 0$. $y= pm 5$ are boundary points of the domain $[-5,5]$ of $h$.
    $endgroup$
    – Fred
    Dec 19 '18 at 13:03











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1 Answer
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1 Answer
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active

oldest

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1












$begingroup$

Your solutions are correct and Lagrange is simpler than polar coordinates.



Another simple method: from $g(x,y)=0$ we get $x^2=25-y^2$, hence $f(x,y)=75-3y^2-y^3$.



Now discuss the function $h(y)=75-3y^2-y^3$ on the intervall $[-5,5]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So I can take the derivative of $75-3y^2-y^3$, set it equal to $0$ and solve for $y$ appropriately. Then solve for $x=pm sqrt{25-y^2}$. I get $-3y(2+y)=0$ which yields again $y=-2$ implies $x= pm sqrt{21}$ and $y=0$ implies $x=pm 5$. But in this case, how do I get the case $x=0, y = pm 5$ ?
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:58






  • 1




    $begingroup$
    The cases $x=0, y= pm 5$ are not obtained with the derivative, since $h'( pm 5) ne 0$. $y= pm 5$ are boundary points of the domain $[-5,5]$ of $h$.
    $endgroup$
    – Fred
    Dec 19 '18 at 13:03
















1












$begingroup$

Your solutions are correct and Lagrange is simpler than polar coordinates.



Another simple method: from $g(x,y)=0$ we get $x^2=25-y^2$, hence $f(x,y)=75-3y^2-y^3$.



Now discuss the function $h(y)=75-3y^2-y^3$ on the intervall $[-5,5]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So I can take the derivative of $75-3y^2-y^3$, set it equal to $0$ and solve for $y$ appropriately. Then solve for $x=pm sqrt{25-y^2}$. I get $-3y(2+y)=0$ which yields again $y=-2$ implies $x= pm sqrt{21}$ and $y=0$ implies $x=pm 5$. But in this case, how do I get the case $x=0, y = pm 5$ ?
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:58






  • 1




    $begingroup$
    The cases $x=0, y= pm 5$ are not obtained with the derivative, since $h'( pm 5) ne 0$. $y= pm 5$ are boundary points of the domain $[-5,5]$ of $h$.
    $endgroup$
    – Fred
    Dec 19 '18 at 13:03














1












1








1





$begingroup$

Your solutions are correct and Lagrange is simpler than polar coordinates.



Another simple method: from $g(x,y)=0$ we get $x^2=25-y^2$, hence $f(x,y)=75-3y^2-y^3$.



Now discuss the function $h(y)=75-3y^2-y^3$ on the intervall $[-5,5]$.






share|cite|improve this answer









$endgroup$



Your solutions are correct and Lagrange is simpler than polar coordinates.



Another simple method: from $g(x,y)=0$ we get $x^2=25-y^2$, hence $f(x,y)=75-3y^2-y^3$.



Now discuss the function $h(y)=75-3y^2-y^3$ on the intervall $[-5,5]$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 12:50









FredFred

44.4k1845




44.4k1845












  • $begingroup$
    So I can take the derivative of $75-3y^2-y^3$, set it equal to $0$ and solve for $y$ appropriately. Then solve for $x=pm sqrt{25-y^2}$. I get $-3y(2+y)=0$ which yields again $y=-2$ implies $x= pm sqrt{21}$ and $y=0$ implies $x=pm 5$. But in this case, how do I get the case $x=0, y = pm 5$ ?
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:58






  • 1




    $begingroup$
    The cases $x=0, y= pm 5$ are not obtained with the derivative, since $h'( pm 5) ne 0$. $y= pm 5$ are boundary points of the domain $[-5,5]$ of $h$.
    $endgroup$
    – Fred
    Dec 19 '18 at 13:03


















  • $begingroup$
    So I can take the derivative of $75-3y^2-y^3$, set it equal to $0$ and solve for $y$ appropriately. Then solve for $x=pm sqrt{25-y^2}$. I get $-3y(2+y)=0$ which yields again $y=-2$ implies $x= pm sqrt{21}$ and $y=0$ implies $x=pm 5$. But in this case, how do I get the case $x=0, y = pm 5$ ?
    $endgroup$
    – Poujh
    Dec 19 '18 at 12:58






  • 1




    $begingroup$
    The cases $x=0, y= pm 5$ are not obtained with the derivative, since $h'( pm 5) ne 0$. $y= pm 5$ are boundary points of the domain $[-5,5]$ of $h$.
    $endgroup$
    – Fred
    Dec 19 '18 at 13:03
















$begingroup$
So I can take the derivative of $75-3y^2-y^3$, set it equal to $0$ and solve for $y$ appropriately. Then solve for $x=pm sqrt{25-y^2}$. I get $-3y(2+y)=0$ which yields again $y=-2$ implies $x= pm sqrt{21}$ and $y=0$ implies $x=pm 5$. But in this case, how do I get the case $x=0, y = pm 5$ ?
$endgroup$
– Poujh
Dec 19 '18 at 12:58




$begingroup$
So I can take the derivative of $75-3y^2-y^3$, set it equal to $0$ and solve for $y$ appropriately. Then solve for $x=pm sqrt{25-y^2}$. I get $-3y(2+y)=0$ which yields again $y=-2$ implies $x= pm sqrt{21}$ and $y=0$ implies $x=pm 5$. But in this case, how do I get the case $x=0, y = pm 5$ ?
$endgroup$
– Poujh
Dec 19 '18 at 12:58




1




1




$begingroup$
The cases $x=0, y= pm 5$ are not obtained with the derivative, since $h'( pm 5) ne 0$. $y= pm 5$ are boundary points of the domain $[-5,5]$ of $h$.
$endgroup$
– Fred
Dec 19 '18 at 13:03




$begingroup$
The cases $x=0, y= pm 5$ are not obtained with the derivative, since $h'( pm 5) ne 0$. $y= pm 5$ are boundary points of the domain $[-5,5]$ of $h$.
$endgroup$
– Fred
Dec 19 '18 at 13:03


















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