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

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

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

extreme points

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

Maximum (\frac{1}{2}, - 1)

Solution steps

Find the critical points:

(\frac{1}{2}, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, - 1) :

Maximum

Maximum (\frac{1}{2}, - 1)

f (x) = 5 x + 5 sin (x)

f (x) = 5 - 3 sin (2 x)

\frac{2 x - 4}{3}

f (x, y) = 8 y^{2} + x^{2} - x^{2} y

f (x) = x^{3} - 3 x, 0 \leq x \leq 2

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