What is the extreme f(x,y)=-5x^2-3y^2+48x-4y+2xy+300 ?

The extreme f(x,y)=-5x^2-3y^2+48x-4y+2xy+300 is Maximum(5,1)

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extreme f(x,y)=-5x^2-3y^2+48x-4y+2xy+300

extreme points

f (x, y) = - 5 x^{2} - 3 y^{2} + 48 x - 4 y + 2 x y + 300

Maximum (5, 1)

Solution steps

Find the critical points:

(5, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = 56

Check the sign of at each critical point

Check critical point

(5, 1) :

Maximum

Maximum (5, 1)

f (x, y) = 400 - 49 x^{2} - 100 y^{2}

f (x) = 3 x^{4} - 8 x^{3} - 144 x^{2} + 1

f (x, y) = - 5 x^{2} - 5 x y - 7 y^{2} + 50 x - 67 y + 9

y = x^{2} + 2 x + 2

\frac{x ^{2}}{x ^{2} + 2}

What is the extreme f(x,y)=-5x^2-3y^2+48x-4y+2xy+300 ?
The extreme f(x,y)=-5x^2-3y^2+48x-4y+2xy+300 is Maximum(5,1)