What is the extreme f(x,y)=(x-4)ln(xy) ?

The extreme f(x,y)=(x-4)ln(xy) is Saddle(4, 1/4)

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extreme f(x,y)=(x-4)ln(xy)

extreme points

f (x, y) = (x - 4) ln (x y)

Saddle (4, \frac{1}{4})

Solution steps

Find the critical points:

(4, \frac{1}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{x - 4}{y ^{2}}

D (x, y) = \frac{- 2 x ^{2} + 16}{x ^{2} y ^{2}}

Check the sign of at each critical point

Check critical point

(4, \frac{1}{4}) :

Saddle

Saddle (4, \frac{1}{4})

f (x) = \frac{1}{2} (7 x - 2), x \leq 3

\frac{1}{t ^{3} + 2}

f (x) = \frac{6}{7} (x^{2} - 9)^{\frac{2}{3}}

f (x, y) = 7 x y + 14 x - x^{2} + 2 y^{2}

4 x^{2} - 8 x^{4}

What is the extreme f(x,y)=(x-4)ln(xy) ?
The extreme f(x,y)=(x-4)ln(xy) is Saddle(4, 1/4)