What is the extreme f(x,y)=4x^2-12x+y^2+2y-10 ?

The extreme f(x,y)=4x^2-12x+y^2+2y-10 is Minimum(3/2 ,-1)

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extreme f(x,y)=4x^2-12x+y^2+2y-10

extreme points

f (x, y) = 4 x^{2} - 12 x + y^{2} + 2 y - 10

Minimum (\frac{3}{2}, - 1)

Solution steps

Find the critical points:

(\frac{3}{2}, - 1)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(\frac{3}{2}, - 1) :

Minimum

Minimum (\frac{3}{2}, - 1)

p (x) = 12 x^{4} + 21 x^{3} + a x^{2} - 21 x - 9

2 x^{4} - 12 x^{2}

- 10 x^{2} + 180 x + 10

f (x) = x x - 3

f (x) = 2 sec (x)

What is the extreme f(x,y)=4x^2-12x+y^2+2y-10 ?
The extreme f(x,y)=4x^2-12x+y^2+2y-10 is Minimum(3/2 ,-1)