What is the nature of the critical points of $f(x,y) = 4x^2 -12xy + 9y^2$.
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Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:
$f(x,y) = 4x^2 -12xy + 9y^2$.
Then,
(1) $f_1 (x,y) = 8x - 12y$
and
(2) $f_2 (x,y) -12x + 18y$.
To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,
$8x - 12y = 0$
$-12x + 18y = 0$.
Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.
Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?
calculus multivariable-calculus
asked Dec 4 at 1:08
K.M
651312
add a comment |
up vote
1
down vote
favorite
Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:
$f(x,y) = 4x^2 -12xy + 9y^2$.
Then,
(1) $f_1 (x,y) = 8x - 12y$
and
(2) $f_2 (x,y) -12x + 18y$.
To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,
$8x - 12y = 0$
$-12x + 18y = 0$.
Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.
Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?
calculus multivariable-calculus
asked Dec 4 at 1:08
K.M
651312
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:
$f(x,y) = 4x^2 -12xy + 9y^2$.
Then,
(1) $f_1 (x,y) = 8x - 12y$
and
(2) $f_2 (x,y) -12x + 18y$.
To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,
$8x - 12y = 0$
$-12x + 18y = 0$.
Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.
Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?
calculus multivariable-calculus
asked Dec 4 at 1:08
K.M
651312
Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:
$f(x,y) = 4x^2 -12xy + 9y^2$.
Then,
(1) $f_1 (x,y) = 8x - 12y$
and
(2) $f_2 (x,y) -12x + 18y$.
To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,
$8x - 12y = 0$
$-12x + 18y = 0$.
Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.
Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?
calculus multivariable-calculus
calculus multivariable-calculus
asked Dec 4 at 1:08
K.M
651312
asked Dec 4 at 1:08
K.M
651312
asked Dec 4 at 1:08
K.M
651312
asked Dec 4 at 1:08
K.M
651312
asked Dec 4 at 1:08
K.M
651312
651312
add a comment |
add a comment |
1 Answer
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up vote
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accepted
$$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$
Hence it attains minimum value when $2x=3y$.
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$
Hence it attains minimum value when $2x=3y$.
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
add a comment |
up vote
3
down vote
accepted
$$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$
Hence it attains minimum value when $2x=3y$.
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$
Hence it attains minimum value when $2x=3y$.
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
$$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$
Hence it attains minimum value when $2x=3y$.
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
answered Dec 4 at 1:14
Siong Thye Goh
97.4k1463116
97.4k1463116
add a comment |
add a comment |
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