What is the extreme f(x,y,λ)=x^2+y^2-(x+2y-5) ?

The extreme f(x,y,λ)=x^2+y^2-(x+2y-5) is Minimum(1/2 ,1)

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extreme f(x,y,λ)=x^2+y^2-(x+2y-5)

extreme points

f (x, y, λ) = x^{2} + y^{2} - (x + 2 y - 5)

Minimum (\frac{1}{2}, 1)

Solution steps

Find the critical points:

(\frac{1}{2}, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, 1) :

Minimum

Minimum (\frac{1}{2}, 1)

f (x) = 7 x^{4} - 28 x^{3 + 7}

(\frac{3}{2})^{x}

f (x) = \frac{x ^{3} - 1}{x ^{3} + 1}

f (x, y) = 2 x^{2} - 5 x + 6 y + 2 y^{2}

f (x) = \frac{( - 3 x ^{3} + 8 x ^{2} - 5 x - 62 )}{x + 2}

What is the extreme f(x,y,λ)=x^2+y^2-(x+2y-5) ?
The extreme f(x,y,λ)=x^2+y^2-(x+2y-5) is Minimum(1/2 ,1)