Mixing
Find the Maximum and Minimum of the Given Function on the Given Plane Region
$begingroup$
I've been good with most of the max/min finding in different regions, but this one's really messing with me. Can anyone lend a hand? Thanks.
z = 2xy
Region is the circular disk $x^2 + y^2 =< 1 $
calculus optimization
asked Dec 6 '14 at 23:51
MockMock
186111
$endgroup$
add a comment |
$begingroup$
I've been good with most of the max/min finding in different regions, but this one's really messing with me. Can anyone lend a hand? Thanks.
z = 2xy
Region is the circular disk $x^2 + y^2 =< 1 $
calculus optimization
asked Dec 6 '14 at 23:51
MockMock
186111
$endgroup$
add a comment |
$begingroup$
I've been good with most of the max/min finding in different regions, but this one's really messing with me. Can anyone lend a hand? Thanks.
z = 2xy
Region is the circular disk $x^2 + y^2 =< 1 $
calculus optimization
asked Dec 6 '14 at 23:51
MockMock
186111
$endgroup$
I've been good with most of the max/min finding in different regions, but this one's really messing with me. Can anyone lend a hand? Thanks.
z = 2xy
Region is the circular disk $x^2 + y^2 =< 1 $
calculus optimization
calculus optimization
asked Dec 6 '14 at 23:51
MockMock
186111
asked Dec 6 '14 at 23:51
MockMock
186111
asked Dec 6 '14 at 23:51
MockMock
186111
asked Dec 6 '14 at 23:51
MockMock
186111
asked Dec 6 '14 at 23:51
MockMock
186111
186111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Max: $ 2 frac{sqrt{2}}{2}frac{sqrt{2}}{2} = 1$.
Min: Negative of the max, by symmetry, hence $-1$.
Hint: Do you know Lagrange multipliers? At an extremum, the gradient of $f(x, y)$ (in your case, that's $(2y, 2x)$) must be a multiple of the gradient of the constraint; in your the constraint function is $g(x, y) = x^2 + y^2 - 1$ so its gradient is $(2x, 2y)$. Hence you need
$$
2y = c 2x \
2x = c 2y
$$
from which you can conclude that $c = 0, 1,$ or $-1$. Then it's all downhill.
You could have an extremum in the interior, where $nabla f(x, y) = 0$, but in this case that happens only at the origin, and $0$ is neither a max nor min for this function. (Check the values at the points of the circle at angles $pm pi/4$.)
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Max: $ 2 frac{sqrt{2}}{2}frac{sqrt{2}}{2} = 1$.
Min: Negative of the max, by symmetry, hence $-1$.
Hint: Do you know Lagrange multipliers? At an extremum, the gradient of $f(x, y)$ (in your case, that's $(2y, 2x)$) must be a multiple of the gradient of the constraint; in your the constraint function is $g(x, y) = x^2 + y^2 - 1$ so its gradient is $(2x, 2y)$. Hence you need
$$
2y = c 2x \
2x = c 2y
$$
from which you can conclude that $c = 0, 1,$ or $-1$. Then it's all downhill.
You could have an extremum in the interior, where $nabla f(x, y) = 0$, but in this case that happens only at the origin, and $0$ is neither a max nor min for this function. (Check the values at the points of the circle at angles $pm pi/4$.)
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
$endgroup$
add a comment |
$begingroup$
Max: $ 2 frac{sqrt{2}}{2}frac{sqrt{2}}{2} = 1$.
Min: Negative of the max, by symmetry, hence $-1$.
Hint: Do you know Lagrange multipliers? At an extremum, the gradient of $f(x, y)$ (in your case, that's $(2y, 2x)$) must be a multiple of the gradient of the constraint; in your the constraint function is $g(x, y) = x^2 + y^2 - 1$ so its gradient is $(2x, 2y)$. Hence you need
$$
2y = c 2x \
2x = c 2y
$$
from which you can conclude that $c = 0, 1,$ or $-1$. Then it's all downhill.
You could have an extremum in the interior, where $nabla f(x, y) = 0$, but in this case that happens only at the origin, and $0$ is neither a max nor min for this function. (Check the values at the points of the circle at angles $pm pi/4$.)
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
$endgroup$
add a comment |
$begingroup$
Max: $ 2 frac{sqrt{2}}{2}frac{sqrt{2}}{2} = 1$.
Min: Negative of the max, by symmetry, hence $-1$.
Hint: Do you know Lagrange multipliers? At an extremum, the gradient of $f(x, y)$ (in your case, that's $(2y, 2x)$) must be a multiple of the gradient of the constraint; in your the constraint function is $g(x, y) = x^2 + y^2 - 1$ so its gradient is $(2x, 2y)$. Hence you need
$$
2y = c 2x \
2x = c 2y
$$
from which you can conclude that $c = 0, 1,$ or $-1$. Then it's all downhill.
You could have an extremum in the interior, where $nabla f(x, y) = 0$, but in this case that happens only at the origin, and $0$ is neither a max nor min for this function. (Check the values at the points of the circle at angles $pm pi/4$.)
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
$endgroup$
Max: $ 2 frac{sqrt{2}}{2}frac{sqrt{2}}{2} = 1$.
Min: Negative of the max, by symmetry, hence $-1$.
Hint: Do you know Lagrange multipliers? At an extremum, the gradient of $f(x, y)$ (in your case, that's $(2y, 2x)$) must be a multiple of the gradient of the constraint; in your the constraint function is $g(x, y) = x^2 + y^2 - 1$ so its gradient is $(2x, 2y)$. Hence you need
$$
2y = c 2x \
2x = c 2y
$$
from which you can conclude that $c = 0, 1,$ or $-1$. Then it's all downhill.
You could have an extremum in the interior, where $nabla f(x, y) = 0$, but in this case that happens only at the origin, and $0$ is neither a max nor min for this function. (Check the values at the points of the circle at angles $pm pi/4$.)
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
answered Dec 6 '14 at 23:58
John HughesJohn Hughes
64.6k24191
64.6k24191
add a comment |
add a comment |
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