How do you find the absolute extrema of f(x,y) = 2x^2 - y^2 + 6y on the disk of radius 4
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How do you find the absolute extrema of f(x,y) = 2x^2 - y^2 + 6y on the disk of radius 4

[From: ] [author: ] [Date: 12-06-26] [Hit: ]
3).The remaining points will be along the boundary, and to find them,h(x,y,lambda) = f(x,......
Use Lagrange Multiplier and 2nd Derivative test.

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).

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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
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