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

The extreme f(x,y)=-8x^2-3y^2+8x-y+2 is Maximum(1/2 ,-1/6)

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extreme f(x,y)=-8x^2-3y^2+8x-y+2

extreme points

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

Maximum (\frac{1}{2}, - \frac{1}{6})

Solution steps

Find the critical points:

(\frac{1}{2}, - \frac{1}{6})

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

D (x, y) = 96

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, - \frac{1}{6}) :

Maximum

Maximum (\frac{1}{2}, - \frac{1}{6})

f (x) = - \frac{13}{x ^{2} - 5 x + 35}

f (x) = \frac{3 x ^{2} - 4 x y + 2 y ^{2}}{4 x + 9 y ^{2}}

(x - 2)^{3} + 3

f (x) = ∣ x + 4 ∣, - 1 \leq x \leq 6

f (x) = - 2 x^{3} + 27 x^{2} - 84 x, 1 \leq x \leq 8

What is the extreme f(x,y)=-8x^2-3y^2+8x-y+2 ?
The extreme f(x,y)=-8x^2-3y^2+8x-y+2 is Maximum(1/2 ,-1/6)