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

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

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

extreme points

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

Maximum (\frac{1}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(\frac{1}{3}, \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 3

D (x, y) = 9

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, \frac{1}{3}) :

Maximum

Maximum (\frac{1}{3}, \frac{1}{3})

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

- x^{2} + 8 x - 9

f (x) = x^{3} - 12 x^{2} - 27 x + 2, 1 \leq x \leq 10

85 y^{35}

2 x^{3} + 9 x^{2} - 24 x

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