What is the extreme T(x,y)=10-2x^2-3y^2 ?

The extreme T(x,y)=10-2x^2-3y^2 is Maximum(0,0)

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extreme T(x,y)=10-2x^2-3y^2

extreme points

T (x, y) = 10 - 2 x^{2} - 3 y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = e^{4 x^{2} + 2 y^{2} - 32 x}

f (x, y) = x y (8 + x) (y - 9)

\frac{3}{2} x^{2} + x y + \frac{1}{108} y^{4}

(3 x - 4)^{2}

\frac{x ^{3}}{x ^{2} + 4}

What is the extreme T(x,y)=10-2x^2-3y^2 ?
The extreme T(x,y)=10-2x^2-3y^2 is Maximum(0,0)