Assignment 30

course Mth 272

M쩢bassignment #030

030.

Applied Calculus II

05-02-2009

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19:10:02

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

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RESPONSE -->

finding the partial deriv

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

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

the critical point is at (0,0) because it's defined for all parts, so it = 4 which is a relative maximum

confidence assessment: 3

You also need to check the mixed partial f_xy to be sure the point isn't a saddle point.

** 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, 0) (be sure to check my mental arithmetic on this one)

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19:10:09

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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RESPONSE -->

answered

confidence assessment: 3

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19:10:18

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

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RESPONSE -->

answered

confidence assessment: 3

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19:10:33

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

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RESPONSE -->

answered

confidence assessment: 3

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19:13:11

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

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RESPONSE -->

fx = 3x^2-6x+3

3(x^2-2x+1)

(x-1)(x-1)

x=1

fy= 3y^2+12y+12

3(y^2+4y+4)

(y+2)(y+2)

y= -2

confidence assessment: 3

** 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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19:14:24

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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RESPONSE -->

f(1,-2) = 0 relative min

becuase x,y > x0,y0

confidence assessment: 3

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19:14:37

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

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RESPONSE -->

answered

confidence assessment: 3

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19:15:04

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

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RESPONSE -->

by factoring and then plugging the values into the original function

confidence assessment: 3

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19:15:34

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

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RESPONSE -->

i didn't use the second partial test. could you explain how i would use it in this case?

confidence assessment: 3

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You don't appear to be considering the mixed partial derivative f_xy, which is part of the test for whether you have a max or a min.

My notes should cover it; let me know if you have questions.