Assignment 30

course Mth 272

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16:58:37

Query problem 7.5.10 extrema of x^2+6xy+10y^2-4y+4

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confidence assessment:

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16:58:56

List the relative extrema and the saddle points of the function and tell which is which, and how you obtained each.

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I understand how to get the fxx and fyy but dont understand how to get the fxy(x,y)

confidence assessment: 0

** fx = 2x + 6y ; fy = 6x + 20 y - 4, where fx and fy mean x and y partial derivatives of f.

fxx is the x derivative of fx and is therefore 2

fyy is the y derivative of fy and is therefore 20

fxy is the y derivative of fx and is therefore 6. Note that fyx is the x derivative of fy and is therefore 6. fxy and fyx are always the same, provided the derivatives exist.

fx = 0 and fy = 0 if

2x + 6y = 0 and

6x + 20y - 4 = 0.

This is a system of two equations in two unknowns. You can solve the system by any of several methods. Multiplying the first equation by -3 and adding the resulting equations, for example, gives you 2 y = 4 so that y = 2. Substituting y = 2 into the first equation gives you 2x + 6 * 2 = 0, which has solution x = 6.

Whatever method you use, the solution of this system is x = -6, y = 2.

To determine the nature of the critical point at (-6, 2) you have to look at fxx, fyy and fxy. We have

fxx = 2

fyy = 20

fxy = 6.

This is a system of two equations in two unknowns. You can solve the system by any of several methods. Multiplying the first equation by -3 and adding the resulting equations, for example, gives you 2 y = 4 so that y = 2. Substituting y = 2 into the first equation gives you 2x + 6 * 2 = 0, which has solution x = -6.

Whatever method you use, the solution of this system is x = -6, y = 2.

To determine the nature of the critical point at (-6, 2) you have to evaluate the quantity fxx * fyy - 4 fxy^2.

We have

fxx = 2

fyy = 20

fxy = 6.

So fxx * fyy - 4 fxy^2 = 2 * 20 - 4 * 6^2 = 8.

This quantity is positive, so you have either a maximum or a minimum.

Since fxx and fyy are both positive the graph is concave upward in all directions and the point is a minimum.

The coordinates of the point are (-6, 2, (-6)^2+6 * (-6) * 2 + 10* 2^2 - 4 * 2 + 4 ) = (-6, 2, 244) (be sure to check my mental arithmetic on this one)

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16:59:16

What are the critical points and what equations did you solve to get them?

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fx(x,y) = 2x + 6y

fy(x,y) = 6x + 20y - 4

2x + 6y = 0

y = -1/3 x

6x - 20/3 x - 4

x = -6

6x + 20y - 4 = 0

x = (4-20y)/6

2(4/20y/6) + 6y = 0

y = 2

critical points are (-6,2)

f(-6,2) = 0

Extrema = (-6,2,0)

confidence assessment: 2

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16:59:29

How did you test each critical point to determine if it is a relative max, relative min or saddle point?

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confidence assessment:

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16:59:31

Query problem 7.5.28 extrema of x^3+y^3 -3x^2+6y^2+3x+12y+7

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confidence assessment:

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16:59:34

List the relative extrema and the saddle points of the function and tell which is which, and how you obtained each.

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confidence assessment:

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16:59:47

What are the critical points and what equations did you solve to get them?

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critical points are (1,-2)

fx(x,y) = 3x^2- 6x + 3 = 3(x-1)(x-1) = 0 when x = 1

fy(x,y) = 3y^2 + 12y + 12 = 3(y+2)(y+s) = 0 when y = -2

f(1,-2) = 0

Extrema is (1,-2,0)

confidence assessment: 2

** We have fx = 3 x^2 + 6 x + 3 and fy = 3 y^2 + 12 y + 12.

Factoring we get

fx = 3 ( x^2 - 2x + 1) = 3 ( x-1)^2 and

fy = 3 ( y^2 + 4y + 4) = 3 ( y+2)^2.

So

fx = 0 when 3(x-1)^2 = 0, or x = 1 and

fy = 0 when 3(y+2)^2 = 0 or y = -2.

We get

fxx = 6 x - 6 (the x derivative of fx)

fyy = 6 y + 12 (the y derivative of fy) and

fxy = 0 (the y derivative of fx, which is also equal to fyx, the x derivative of fy)

At the critical point x = 1, y = -2 we get fxx = fyy = 0. So the test for max, min and critical point gives us fxx*fyy - 4 fxy^2 = 0, which is inconclusive--it tells us nothing about max, min or saddle point. **

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16:59:50

How did you test each critical point to determine if it is a relative max, relative min or saddle point?

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confidence assessment:

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16:59:53

At what point(s) did the second-partials test fail?

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confidence assessment:

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Good work, except for the parts requiring the 'mixed' derivatives. See my notes and let me know if they don't clarify that.