What is the extreme f(x,y)=5x^2-4xy+4y^2+12x+25 ?

The extreme f(x,y)=5x^2-4xy+4y^2+12x+25 is Minimum(-3/2 ,-3/4)

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extreme f(x,y)=5x^2-4xy+4y^2+12x+25

extreme points

f (x, y) = 5 x^{2} - 4 x y + 4 y^{2} + 12 x + 25

Minimum (- \frac{3}{2}, - \frac{3}{4})

Solution steps

Find the critical points:

(- \frac{3}{2}, - \frac{3}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 64

Check the sign of at each critical point

Check critical point

(- \frac{3}{2}, - \frac{3}{4}) :

Minimum

Minimum (- \frac{3}{2}, - \frac{3}{4})

f (x) = 2 x^{3} - \frac{12 x ^{2}}{2} - 2 [- 1.3]

\frac{27 x ^{2}}{( 2 - x ) ^{3}}

\frac{8 x ^{2}}{x - 6}, - 3 \leq x \leq 2

f (x, y) = x + 3 y

g (x) = x^{3} - 12 x^{2} + 45 x + 4

What is the extreme f(x,y)=5x^2-4xy+4y^2+12x+25 ?
The extreme f(x,y)=5x^2-4xy+4y^2+12x+25 is Minimum(-3/2 ,-3/4)