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# Using Lagrange multipliers, what is the absolute extrema of f(x,y) =x^2 +y subject to the constraint x^2+y^2 = 9?

Update:

I tried doing the F(x,y,λ) = f(x,y) + λg(x,y). Then i get the ∂/∂x ,∂/∂y and ∂/∂λ. I got an answer but it said I was wrong

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- VamanLv 72 months ago
I tried doing the F(x,y,λ) = f(x,y) + λg(x,y). Then i get the ∂/∂x ,∂/∂y and ∂/∂λ. I got an answer but it said I was wrong. Your aim is to find lambda which I write as l. Write g(x,y)= x^2+y +l(x^2+y^2-9). Now take the partial derivative wrt x and y and equate them to 0.

dg/dx= 2x +2lx=0. x is not 0. This gives l=-1.

dg/dy= 1+2ly=0. y=1/2,. To find x use x^2+y^2=9

x^2=9-1/4= 35/4. x= +/- sqrt (35)/2. Take.

x= + sqrt(35/4), f(x,y) =35/4+1/2=37/4. It seems that f(x,y) has only a maximum.

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