Global extrema of $f(x,y)=3x^2-y^3$ on the circle $x^2+y^2 leq 25$
$begingroup$
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
asked Dec 19 '18 at 12:41
PoujhPoujh
569516
$endgroup$
add a comment |
$begingroup$
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
asked Dec 19 '18 at 12:41
PoujhPoujh
569516
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 19 '18 at 12:41
PoujhPoujh
569516
$endgroup$
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
asked Dec 19 '18 at 12:41
PoujhPoujh
569516
asked Dec 19 '18 at 12:41
PoujhPoujh
569516
edited Jan 7 at 19:20
Poujh
asked Dec 19 '18 at 12:41
PoujhPoujh
569516
asked Dec 19 '18 at 12:41
PoujhPoujh
569516
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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 $[-5,5]$.
answered Dec 19 '18 at 12:50
FredFred
44.4k1845
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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 $[-5,5]$.
answered Dec 19 '18 at 12:50
FredFred
44.4k1845
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $[-5,5]$.
answered Dec 19 '18 at 12:50
FredFred
44.4k1845
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $[-5,5]$.
answered Dec 19 '18 at 12:50
FredFred
44.4k1845
$endgroup$
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 $[-5,5]$.
answered Dec 19 '18 at 12:50
FredFred
44.4k1845
answered Dec 19 '18 at 12:50
FredFred
44.4k1845
answered Dec 19 '18 at 12:50
FredFred
44.4k1845
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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