7-5  Assignment 30

course Mth 272

̦n󛠈}~nassignment #030

030.

Applied Calculus II

09-08-2008

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00:08:30

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RESPONSE -->

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00:13:35

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

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RESPONSE -->

fx = 2x + 6y

Solve for x

x = -3y

fy = 6x + 20y - 4

Use x = 3y to evaluate

6(-3y) + 20y - 4

-18y + 20y - 4

2y = 4

y = 2

Then Evaluate for x

x = -3(2)

x = -6

So...(-6)^2+6(-6)(2)+10(2)^2-4(2)+4

36 - 72 + 40 - 8 + 4

0

Therefore... (-6,2,0)

confidence assessment: 2

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00:13:42

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 -->

fx = 2x + 6y

Solve for x

x = -3y

fy = 6x + 20y - 4

Use x = 3y to evaluate

6(-3y) + 20y - 4

-18y + 20y - 4

2y = 4

y = 2

Then Evaluate for x

x = -3(2)

x = -6

So...(-6)^2+6(-6)(2)+10(2)^2-4(2)+4

36 - 72 + 40 - 8 + 4

0

Therefore... (-6,2,0)

confidence assessment: 2

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00:13:49

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

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RESPONSE -->

fx = 2x + 6y

Solve for x

x = -3y

fy = 6x + 20y - 4

Use x = 3y to evaluate

6(-3y) + 20y - 4

-18y + 20y - 4

2y = 4

y = 2

Then Evaluate for x

x = -3(2)

x = -6

So...(-6)^2+6(-6)(2)+10(2)^2-4(2)+4

36 - 72 + 40 - 8 + 4

0

Therefore... (-6,2,0)

confidence assessment: 2

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00:13:59

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 -->

fx = 2x + 6y

Solve for x

x = -3y

fy = 6x + 20y - 4

Use x = 3y to evaluate

6(-3y) + 20y - 4

-18y + 20y - 4

2y = 4

y = 2

Then Evaluate for x

x = -3(2)

x = -6

So...(-6)^2+6(-6)(2)+10(2)^2-4(2)+4

36 - 72 + 40 - 8 + 4

0

Therefore... (-6,2,0)

confidence assessment: 1

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00:17:47

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)

= 3 (x-1)(x-1)

Solve for x....x = 1

fy= 3y^2 + 12y + 12

= 3(y^2 + 6y + 6)

should be

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

= y(y+6) = -6

y + 6 = -6

y = 0

(1)^3+(0)^3 -3(1)^2+6(0)^2+3(1)+12(0)+7

1 + 0 + 3 + 0 + 3 + 0 + 7

4 + 3 + 7

14

(1, 0, 14)

confidence assessment: 2

You have just one small arithmetic error; your procedure is fine. See the note I inserted above. Compare with the solution given below, which I believe is correct:

** 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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00:17:53

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 -->

fx = 3x^2 - 6x + 3

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

= 3 (x-1)(x-1)

Solve for x....x = 1

fy= 3y^2 + 12y + 12

= 3(y^2 + 6y + 6)

= y(y+6) = -6

y + 6 = -6

y = 0

(1)^3+(0)^3 -3(1)^2+6(0)^2+3(1)+12(0)+7

1 + 0 + 3 + 0 + 3 + 0 + 7

4 + 3 + 7

14

(1, 0, 14)

confidence assessment: 2

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00:17:58

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

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RESPONSE -->

fx = 3x^2 - 6x + 3

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

= 3 (x-1)(x-1)

Solve for x....x = 1

fy= 3y^2 + 12y + 12

= 3(y^2 + 6y + 6)

= y(y+6) = -6

y + 6 = -6

y = 0

(1)^3+(0)^3 -3(1)^2+6(0)^2+3(1)+12(0)+7

1 + 0 + 3 + 0 + 3 + 0 + 7

4 + 3 + 7

14

(1, 0, 14)

confidence assessment: 1

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00:18:06

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 -->

fx = 3x^2 - 6x + 3

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

= 3 (x-1)(x-1)

Solve for x....x = 1

fy= 3y^2 + 12y + 12

= 3(y^2 + 6y + 6)

= y(y+6) = -6

y + 6 = -6

y = 0

(1)^3+(0)^3 -3(1)^2+6(0)^2+3(1)+12(0)+7

1 + 0 + 3 + 0 + 3 + 0 + 7

4 + 3 + 7

14

(1, 0, 14)

confidence assessment: 2

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00:18:22

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

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RESPONSE -->

fx = 3x^2 - 6x + 3

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

= 3 (x-1)(x-1)

Solve for x....x = 1

fy= 3y^2 + 12y + 12

= 3(y^2 + 6y + 6)

= y(y+6) = -6

y + 6 = -6

y = 0

confidence assessment: 1

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You are in good shape here. You made one minor error in your arithmetic, which affected your answer to one problem, but everything you are doing appears to be correct.