course Mth 272
Please see the note I included in the last problem. Thanks.
¹âöîËŸäÀw§€´»³y‚±ÔÓ玢assignment #030
030.
Applied Calculus II
11-26-2008
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14:32:01
**** Query problem 7.5.10 extrema of x^2+6xy+10y^2-4y+4
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+6=0
x= -3
fy 20y+2= 0
y= 0.1
confidence assessment: 2
** 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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14:32:59
What are the critical points and what equations did you solve to get them?
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RESPONSE -->
-3^2 +6(-3)(-.1) + 10(-0.1)^2 - 4(-0.1) + 4
= (-3,-0.1, 15.3)
confidence assessment: 2
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14:36:20
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
fy= 3y^2 + 12y +12
3x^2-6x= -3
x^2-2x= -1
3y^2 +12y= -12
y^2 + 4y= -4
Professor Smith,
I am confused on this problem. Do I need to use the quadratic equation to solve for x,y or am I on the wrong track? It's been a while since I have used the quadratic eqn which is why I cannot help but believe I've already done something wrong in the problem. I know I need to solve for x and y but I'm wondering if that is the right way. Please let me know and I can complete the problem. Thanks.
confidence assessment: 0
Generally you solve a quadratic equation by putting it into the form a x^2 + b x + c = 0 and using the quadratic formula. In the present problem the partial derivatives are squares of binomials. You should review the solution of quadratic equations in the review chapter at the beginning of the text; online resources of your choice can also be useful. Just search for 'solving quadratic equations' or something of this nature.
For the present problem:
** 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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&#Your work looks good. See my notes. Let me know if you have any questions. &#