What is the extreme f(xy)=2x^2+3/2 y^2+3xy-10x-9y+2 ?

The extreme f(xy)=2x^2+3/2 y^2+3xy-10x-9y+2 is Minimum(1,2)

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extreme f(xy)=2x^2+3/2 y^2+3xy-10x-9y+2

extreme points

f (x y) = 2 x^{2} + \frac{3}{2} y^{2} + 3 x y - 10 x - 9 y + 2

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 3

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

2 x + 1 - 2 x

f (x) = y = \frac{3 x ^{2}}{x + 1}, - 1 \leq x \leq 5

2 x^{3} - 12 x^{2} + 18 x

f (x) = (- x^{x} (ln (x) + 1)^{2} - x^{x - 1})

f (x) = \frac{2 e ^{- 3 x}}{2 - 3 x}

What is the extreme f(xy)=2x^2+3/2 y^2+3xy-10x-9y+2 ?
The extreme f(xy)=2x^2+3/2 y^2+3xy-10x-9y+2 is Minimum(1,2)