What is the extreme f(x,y)=-3x^2-4xy-8y^2+42x+68y+4 ?

The extreme f(x,y)=-3x^2-4xy-8y^2+42x+68y+4 is Maximum(5,3)

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extreme f(x,y)=-3x^2-4xy-8y^2+42x+68y+4

extreme points

f (x, y) = - 3 x^{2} - 4 x y - 8 y^{2} + 42 x + 68 y + 4

Maximum (5, 3)

Solution steps

Find the critical points:

(5, 3)

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

D (x, y) = 80

Check the sign of at each critical point

Check critical point

(5, 3) :

Maximum

Maximum (5, 3)

f (x) = \frac{1}{3} x^{3} - x^{2} + 3 x + 4

f (x) = 2 (2 x - 8)

f (x) = - x^{2} - 4 x - 9

- 47 x^{2} + 46.7 x + 68296.6

f (x) = \frac{x}{x ^{2} + 16}, 0 \leq x \leq 8

What is the extreme f(x,y)=-3x^2-4xy-8y^2+42x+68y+4 ?
The extreme f(x,y)=-3x^2-4xy-8y^2+42x+68y+4 is Maximum(5,3)