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I'm struggling with this question below...?
How could you find a point in the x-y plane where the function listed has a local minimum.
f(x, y) = 4x^2 + 3xy + 5y^2 + x + y + 1
If there is such a point but it is not listed, choose answer I. If there is none, answer J.
these are the solutions to choose from in the back of the book...
a. (-9/79, -7/79)
b. (-2/19, -3/38)
c. (-7/71, -5/71)
d. (-3/32, -1/16)
e. (-1/11, -3/55)
f. (-1/11, -1/22)
g. (-3/31, -1/31)
h. (-1/8, 0)
i. There is a local minimum but it is not listed.
j. There is no local minimum for this surface.
3 Answers
- Anonymous1 decade agoFavorite Answer
df/dx = 8x + 3y + 1
df/dy = 3x + 10y + 1
Set each = zero and solve:
from top: y = -(8x + 1)/3
Put into bottom:
3x -(10/3)(8x+1) = -1
solve for x:
-71/3x = 7/3
x = -7/71.
So the front-runner is c. You need to now solve for y, though.
;-)
Source(s): Oh yeah, don't forget to check to make sure it's actually a min (which it is, I can see), and not a max. I think that's a generalized second-derivative test. (Jacobian? Hessian?, ugh, time to brush up on this stuff . . . ) - santmann2002Lv 71 decade ago
df/dx=8x+3y+1=0
df/dy=3x+10y+1=0
solving
24x+9y+3=0
24x+80y +8=0 71y+5=0 so y = -5/71
8x= 15/71-71/71 x= -7/71
The only critical point is (-7/71,-5/71)
To find out what happens at this point we must calculate the Hessian.
fxx=8
fyy=10
fxy= 3
fxx*fyy-fxy^2= 71>0 .As fxx>0 it is a local minimum (c)
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