Use Lagrange Multiplier and 2nd Derivative test.
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Thanks! :)
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The maxima are at (+/-sqrt(15),1), and the minimum is at (0,-4). To show this, read on:
We want to find the extreme values of
f(x,y) = 2x^2 - y^2 + 6y
constrained to the disk of radius 4. The boundary of this region is given by
g(x,y) = x^2+y^2 - 16 = 0.
The first thing to do is to find the interior critical points. This is done by simultaneously solving for
df/dx = 4x = 0
df/dy = -2y+6 = 0
==> (0,3).
The remaining points will be along the boundary, and to find them, we use the Lagrange multiplier lambda to define
h(x,y,lambda) = f(x,y) + lambda*g(x,y).
The critical points are now defined by
dh/dx = 4x + lambda*2*x = 0 ==> 2x(lambda+2) = 0
dh/dy = -2y+6 + lambda*2*y = 0 ==> 2y(lambda-1) + 6 = 0
dh/dlambda = x^2+y^2 - 16= 0 ==> x^2+y^2 = 16
The second equation tells us that y=3/(1-lambda), provided lambda is not equal to 1.
The first equation tells us that lambda = -2, or x=0. If we first assume that lambda=-2, we have that
y = 1,
g(x,1) = x^2 + 1^2 = 16
==> (sqrt(15),1), (-sqrt(15),1)
However, if instead we assume that x=0, we don't need to explicitly obtain lambda, but instead find that
g(0,y) = 0^2 + y^2 = 16
==> (0,4), (0,-4)
So to summarize, the four extrema on the disk of radius 4 are:
(sqrt(15),1), (-sqrt(15),1), (0,4), (0,-4)
and the interior extrema is
(0,3).
We now need to find which of these values are the largest and smallest; i.e., the absolute extrema. The interior point can be classified using the second derivatives test
f_xx = 4
f_yy = -2
f_xy = 0
D = f_xx f_yy - (f_xy)^2 = -8.
Since, D<0, the point will be a saddle point, and cannot be an extremal value of f(x,y). Thus, we check the boundary points.
f(+/-sqrt(15),1) = 35
f(0,+/-4) = -16+/-24 = -40,8
So, the maxima are at (+/-sqrt(15),1), and the minimum is at (0,-4).
We want to find the extreme values of
f(x,y) = 2x^2 - y^2 + 6y
constrained to the disk of radius 4. The boundary of this region is given by
g(x,y) = x^2+y^2 - 16 = 0.
The first thing to do is to find the interior critical points. This is done by simultaneously solving for
df/dx = 4x = 0
df/dy = -2y+6 = 0
==> (0,3).
The remaining points will be along the boundary, and to find them, we use the Lagrange multiplier lambda to define
h(x,y,lambda) = f(x,y) + lambda*g(x,y).
The critical points are now defined by
dh/dx = 4x + lambda*2*x = 0 ==> 2x(lambda+2) = 0
dh/dy = -2y+6 + lambda*2*y = 0 ==> 2y(lambda-1) + 6 = 0
dh/dlambda = x^2+y^2 - 16= 0 ==> x^2+y^2 = 16
The second equation tells us that y=3/(1-lambda), provided lambda is not equal to 1.
The first equation tells us that lambda = -2, or x=0. If we first assume that lambda=-2, we have that
y = 1,
g(x,1) = x^2 + 1^2 = 16
==> (sqrt(15),1), (-sqrt(15),1)
However, if instead we assume that x=0, we don't need to explicitly obtain lambda, but instead find that
g(0,y) = 0^2 + y^2 = 16
==> (0,4), (0,-4)
So to summarize, the four extrema on the disk of radius 4 are:
(sqrt(15),1), (-sqrt(15),1), (0,4), (0,-4)
and the interior extrema is
(0,3).
We now need to find which of these values are the largest and smallest; i.e., the absolute extrema. The interior point can be classified using the second derivatives test
f_xx = 4
f_yy = -2
f_xy = 0
D = f_xx f_yy - (f_xy)^2 = -8.
Since, D<0, the point will be a saddle point, and cannot be an extremal value of f(x,y). Thus, we check the boundary points.
f(+/-sqrt(15),1) = 35
f(0,+/-4) = -16+/-24 = -40,8
So, the maxima are at (+/-sqrt(15),1), and the minimum is at (0,-4).
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Lagrange Multiplier method,
F(x,y) = 2x^2 - y^2 + 6y - k(x^2 + y^2 - 4^2)
∂F/∂x = 4x - 2 k x = 0. . . . . . . . . . (equation 1)
∂F/∂y = -2 (k y + y - 3) = 0. . . . . . . . . . (equation 2)
∂F/∂k = -x^2 - y^2 + 16 = 0. . . . . . . . . . (equation 3)
solve simultaneous equation above and we get,
k = -7/4, x = 0, y = -4
k = -1/4, x = 0, y = 4
k = 2, x = -√15, y = 1
k = 2, x = √15, y = 1
f(x,y) = 2(0)^2 - (-4)^2 + 6(-4) = -40
f(x,y) = 2(0)^2 - (4)^2 + 6(4) = 8
f(x,y) = 2(-√15)^2 - (1)^2 + 6(1) = 35
f(x,y) = 2(√15)^2 - (1)^2 + 6(1) = 35
F(x,y) = 2x^2 - y^2 + 6y - k(x^2 + y^2 - 4^2)
∂F/∂x = 4x - 2 k x = 0. . . . . . . . . . (equation 1)
∂F/∂y = -2 (k y + y - 3) = 0. . . . . . . . . . (equation 2)
∂F/∂k = -x^2 - y^2 + 16 = 0. . . . . . . . . . (equation 3)
solve simultaneous equation above and we get,
k = -7/4, x = 0, y = -4
k = -1/4, x = 0, y = 4
k = 2, x = -√15, y = 1
k = 2, x = √15, y = 1
f(x,y) = 2(0)^2 - (-4)^2 + 6(-4) = -40
f(x,y) = 2(0)^2 - (4)^2 + 6(4) = 8
f(x,y) = 2(-√15)^2 - (1)^2 + 6(1) = 35
f(x,y) = 2(√15)^2 - (1)^2 + 6(1) = 35