What is the extreme f(x,y)=x^2+4xy+5y^2-7x+9y-10 ?

The extreme f(x,y)=x^2+4xy+5y^2-7x+9y-10 is Minimum(53/2 ,-23/2)

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extreme f(x,y)=x^2+4xy+5y^2-7x+9y-10

extreme points

f (x, y) = x^{2} + 4 x y + 5 y^{2} - 7 x + 9 y - 10

Minimum (\frac{53}{2}, - \frac{23}{2})

Solution steps

Find the critical points:

(\frac{53}{2}, - \frac{23}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(\frac{53}{2}, - \frac{23}{2}) :

Minimum

Minimum (\frac{53}{2}, - \frac{23}{2})

f (x) = 9 x - 18 cos (x), - 2 \leq x \leq 0

f (x) = 2 x^{2} - 4 x, 0 \leq x \leq 3

x^{2} - 18 x + 79

f (x) = 3 x^{4} - 294 x^{2} + 3, - 8 \leq x \leq 8

f (x) = 2 x^{2} - 7, - 7 \leq x \leq 7

What is the extreme f(x,y)=x^2+4xy+5y^2-7x+9y-10 ?
The extreme f(x,y)=x^2+4xy+5y^2-7x+9y-10 is Minimum(53/2 ,-23/2)