What is the extreme f(x,y)=x^2+y^2-20x+4y-10 ?

The extreme f(x,y)=x^2+y^2-20x+4y-10 is Minimum(10,-2)

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extreme f(x,y)=x^2+y^2-20x+4y-10

extreme points

f (x, y) = x^{2} + y^{2} - 20 x + 4 y - 10

Minimum (10, - 2)

Solution steps

Find the critical points:

(10, - 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(10, - 2) :

Minimum

Minimum (10, - 2)

2 x^{3} + 6 x^{2} - 90 x + 7, - 5 \leq x \leq 6

f (x) = - \frac{5}{4} x^{2} + \frac{5}{2} x + \frac{7}{4}

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

f (x) = e^{2 x} + e^{(- x)}

- x^{2} + 8 x - 9

What is the extreme f(x,y)=x^2+y^2-20x+4y-10 ?
The extreme f(x,y)=x^2+y^2-20x+4y-10 is Minimum(10,-2)