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

The extreme f(x,y)=2x^2+y^2-3x is Minimum(3/4 ,0)

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

extreme points

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

Minimum (\frac{3}{4}, 0)

Solution steps

Find the critical points:

(\frac{3}{4}, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(\frac{3}{4}, 0) :

Minimum

Minimum (\frac{3}{4}, 0)

f (x) = - (x + 4)^{3}, - 4 \leq x \leq - 3

f (x) = \frac{x ^{3}}{3} - \frac{7 x ^{2}}{2} + 10 x - 4

f (x) = \frac{4 x ^{2}}{x ^{2} - 4}

f (x) = x^{2} + 2 x + y^{2}

f (x, y) = 4 - x^{2} - y^{2}

What is the extreme f(x,y)=2x^2+y^2-3x ?
The extreme f(x,y)=2x^2+y^2-3x is Minimum(3/4 ,0)