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spndsh Group TitleBest ResponseYou've already chosen the best response.1
dw:1352550338176:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
you last equation does not look correct
 one year ago

spndsh Group TitleBest ResponseYou've already chosen the best response.1
dw:1352551028612:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
I would solve for lambda using the 1st eq. plug into the 2nd equation and solve for x (or y) and finally use the last equation...
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
btw, they are all set = to 0, and you should show this...
 one year ago

spndsh Group TitleBest ResponseYou've already chosen the best response.1
would that give the extreme points?
 one year ago

spndsh Group TitleBest ResponseYou've already chosen the best response.1
dw:1352552076296:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
I think you missed a factor of 9 on the y^2 term
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
but the crucial step is in solving, you should have found y=0 as a possible root
 one year ago

spndsh Group TitleBest ResponseYou've already chosen the best response.1
dw:1352553005461:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
no. you can do with z= y^2 4z^2 +9z9=0
 one year ago

spndsh Group TitleBest ResponseYou've already chosen the best response.1
ok, whats the next step?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
\[ 6x^2 2 \lambda x=0 > \lambda= 3x\] \[ 4y^3 2\lambda y=0 > y=0 \text{ or } 2y^23x=0 \] notice y = 0 is a possible solution \[ y^2= \frac{3}{2} x \] using y^2 = 3/2 x in \[ x^2 + y^2 1=0 > x^2 +\frac{3}{2}x 1 =0\] \[ x= \frac{1}{4}(3 ± 5) \] when we plug in for the various candidates (1,0) is the max value x= 2 is outside the region, x=1/2 is an inflection point
 one year ago

spndsh Group TitleBest ResponseYou've already chosen the best response.1
ok, thanks so much
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
After looking at this a little more, we have possible solutions at x=0, y=±1 y=0, x=±1 x= 1/2, y= ± 3√/2 the value of f(x,y) at these points is (0,1) 1 (0,1) 1 (1,0) 1 (1,0) 2 (1/2, sqrt(3)/2) 13/16 (1/2, sqrt(3)/2) 13/16 so point (1,0) is a max with value 2 point(1,0) is a min with value 1 points (1/2,±sqrt(3)/2) is a local min points (0,±1) are local maximums
 one year ago
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