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

The extreme T(x,y)=ln(3xy+2x^2-y) is Saddle(1/3 ,-4/9)

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extreme T(x,y)=ln(3xy+2x^2-y)

extreme points

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

Saddle (\frac{1}{3}, - \frac{4}{9})

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{4}{9})

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

D (x, y) = \frac{36 x ^{4} + 108 x ^{3} y + 81 x ^{2} y ^{2} - 36 x ^{2} y - 8 x ^{2} - 54 x y ^{2} - 12 x y + 9 y ^{2} + 4 y}{( - y + 2 x ^{2} + 3 x y ) ^{4}}

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - \frac{4}{9}) :

Saddle

Saddle (\frac{1}{3}, - \frac{4}{9})

f (x) = 5000 - 15 x + 0.08 x^{2}

2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

f (x) = - 2 x^{2} + 200 x

y = x^{2} - 5 x - 14

x \cdot 7 - x

What is the extreme T(x,y)=ln(3xy+2x^2-y) ?
The extreme T(x,y)=ln(3xy+2x^2-y) is Saddle(1/3 ,-4/9)