What is the extreme f(x,y)=x^2+1/2 y^2+1/2 (y-x)-3/2 ?

The extreme f(x,y)=x^2+1/2 y^2+1/2 (y-x)-3/2 is Minimum(1/4 ,-1/2)

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extreme f(x,y)=x^2+1/2 y^2+1/2 (y-x)-3/2

extreme points

f (x, y) = x^{2} + \frac{1}{2} y^{2} + \frac{1}{2} (y - x) - \frac{3}{2}

Minimum (\frac{1}{4}, - \frac{1}{2})

Solution steps

Find the critical points:

(\frac{1}{4}, - \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 1

D (x, y) = 2

Check the sign of at each critical point

Check critical point

(\frac{1}{4}, - \frac{1}{2}) :

Minimum

Minimum (\frac{1}{4}, - \frac{1}{2})

x^{4} + 3 x^{3} + 2 x^{2} + x + 1

f (x) = - 0.4 x^{2} + 90 x - 2000, 0 \leq x

2 x^{2} - 2 x

f (x) = (x - 2)^{\frac{1}{8}}

y = 2 cos (x) - 11 x + 7 [- π, 0]

What is the extreme f(x,y)=x^2+1/2 y^2+1/2 (y-x)-3/2 ?
The extreme f(x,y)=x^2+1/2 y^2+1/2 (y-x)-3/2 is Minimum(1/4 ,-1/2)