What is the extreme f(x,y)=7x^2+8y^2+4x-5y+8 ?

The extreme f(x,y)=7x^2+8y^2+4x-5y+8 is Minimum(-2/7 , 5/16)

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extreme f(x,y)=7x^2+8y^2+4x-5y+8

extreme points

f (x, y) = 7 x^{2} + 8 y^{2} + 4 x - 5 y + 8

Minimum (- \frac{2}{7}, \frac{5}{16})

Solution steps

Find the critical points:

(- \frac{2}{7}, \frac{5}{16})

\frac{\partial ^{2} f}{\partial x ^{2}} = 16

D (x, y) = 224

Check the sign of at each critical point

Check critical point

(- \frac{2}{7}, \frac{5}{16}) :

Minimum

Minimum (- \frac{2}{7}, \frac{5}{16})

f (x) = - 7 sec (x), - \frac{π}{3} \leq x \leq \frac{π}{3}

f (x, y) = \frac{x y ( 9 - x - 2 y )}{3}

x^{3} + 3 x y^{2} - 3 y^{2} - 3 x^{2} + 11

0.001 x^{2} + 3.4 x - 90

f (x) = 12 x - 12 y - 2 x^{2} - 3 y^{2} - 25

What is the extreme f(x,y)=7x^2+8y^2+4x-5y+8 ?
The extreme f(x,y)=7x^2+8y^2+4x-5y+8 is Minimum(-2/7 , 5/16)