What is the extreme f(x)=-2x^2-5xy-6y^2+32x+63y+9 ?

The extreme f(x)=-2x^2-5xy-6y^2+32x+63y+9 is Maximum(3,4)

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extreme f(x)=-2x^2-5xy-6y^2+32x+63y+9

extreme points

f (x) = - 2 x^{2} - 5 x y - 6 y^{2} + 32 x + 63 y + 9

Maximum (3, 4)

Solution steps

Find the critical points:

(3, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 12

D (x, y) = 23

Check the sign of at each critical point

Check critical point

(3, 4) :

Maximum

Maximum (3, 4)

f (x) = 4 sin (x) - 3 cos (x)

y = (\frac{I x + 2 I - 1}{2 x + 3}) e^{2 x}

x + \frac{3}{2} x^{\frac{2}{3}}

+ ∣ x^{2} - 3 x + 2 ∣

f (x) = (- 5 x^{2} + 10 x y - 20 x - 7 y^{2} + 240 y - 5300)

What is the extreme f(x)=-2x^2-5xy-6y^2+32x+63y+9 ?
The extreme f(x)=-2x^2-5xy-6y^2+32x+63y+9 is Maximum(3,4)