What is the extreme f(x*y)=3x^2-4xy+3y^2+8x-17y+5 ?

The extreme f(x*y)=3x^2-4xy+3y^2+8x-17y+5 is Minimum(1, 7/2)

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extreme f(x*y)=3x^2-4xy+3y^2+8x-17y+5

extreme points

f (x \cdot y) = 3 x^{2} - 4 x y + 3 y^{2} + 8 x - 17 y + 5

Minimum (1, \frac{7}{2})

Solution steps

Find the critical points:

(1, \frac{7}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 20

Check the sign of at each critical point

Check critical point

(1, \frac{7}{2}) :

Minimum

Minimum (1, \frac{7}{2})

x^{3} - 16 x y + 4 y^{2} + 16 x^{2} - 12 x + 5

f (x) = 6 x^{3} + 2 x

y = x^{3} - 3 x^{2} + 4 x - 4

f (x y) = x y

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 18

What is the extreme f(x*y)=3x^2-4xy+3y^2+8x-17y+5 ?
The extreme f(x*y)=3x^2-4xy+3y^2+8x-17y+5 is Minimum(1, 7/2)