What is the extreme f(x,y)=ln(x)+ln(y)-ln(x+y)-x/(12)-y/3 ?

The extreme f(x,y)=ln(x)+ln(y)-ln(x+y)-x/(12)-y/3 is Saddle(-12,6),Maximum(4,2)

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extreme f(x,y)=ln(x)+ln(y)-ln(x+y)-x/(12)-y/3

extreme points

f (x, y) = ln (x) + ln (y) - ln (x + y) - \frac{x}{12} - \frac{y}{3}

Saddle (- 12, 6), Maximum (4, 2)

Solution steps

Find the critical points:

(- 12, 6), (4, 2)

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

D (x, y) = \frac{2}{x y ( x + y ) ^{2}}

Check the sign of at each critical point

Check critical point

(- 12, 6) :

Saddle

Check critical point

(4, 2) :

Maximum

Saddle (- 12, 6), Maximum (4, 2)

f (x, y) = 8 x^{4} + y^{2} - 16 x y

f (x) = - 2 x - 1

f (x, y) = \frac{x y}{x + y}

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

f (x, y) = e^{- y} (x^{2} + y^{2}) + 6

What is the extreme f(x,y)=ln(x)+ln(y)-ln(x+y)-x/(12)-y/3 ?
The extreme f(x,y)=ln(x)+ln(y)-ln(x+y)-x/(12)-y/3 is Saddle(-12,6),Maximum(4,2)