What is the extreme f(x,y)=x^2+y^2+xy-8x-7y+400 ?

The extreme f(x,y)=x^2+y^2+xy-8x-7y+400 is Minimum(3,2)

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extreme f(x,y)=x^2+y^2+xy-8x-7y+400

extreme points

f (x, y) = x^{2} + y^{2} + x y - 8 x - 7 y + 400

Minimum (3, 2)

Solution steps

Find the critical points:

(3, 2)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(3, 2) :

Minimum

Minimum (3, 2)

f (x) = \frac{( x ^{2} - 12 )}{x - 4}

f (t) = t e^{\frac{w}{t}}

f (x) = y^{2} - x y - x^{2}

f (x) = - x^{2} + 4 x + 5

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

What is the extreme f(x,y)=x^2+y^2+xy-8x-7y+400 ?
The extreme f(x,y)=x^2+y^2+xy-8x-7y+400 is Minimum(3,2)