What is the extreme f(x,y)=1-x^2-x-3y^2+y ?

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

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extreme f(x,y)=1-x^2-x-3y^2+y

extreme points

f (x, y) = 1 - x^{2} - x - 3 y^{2} + y

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) = 12

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) = 1.4 t e^{- 2.9 t}

f (x) = 2 x^{3} - 9 x^{2} - 24 x + 9, - 2 \leq x \leq 5

f (x) = \frac{9 x}{x - 4}

f (x) = x e^{\frac{- x ^{2}}{32}}

2400 = 30 x \cdot 10 y

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