What is the extreme f(x,y)= 1/x-(64)/y+xy ?

The extreme f(x,y)= 1/x-(64)/y+xy is Maximum(-1/4 ,16)

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extreme f(x,y)= 1/x-(64)/y+xy

extreme points

f (x, y) = \frac{1}{x} - \frac{64}{y} + x y

Maximum (- \frac{1}{4}, 16)

Solution steps

Find the critical points:

(- \frac{1}{4}, 16)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{128}{y ^{3}}

D (x, y) = - \frac{256}{x ^{3} y ^{3}} - 1

Check the sign of at each critical point

Check critical point

(- \frac{1}{4}, 16) :

Maximum

Maximum (- \frac{1}{4}, 16)

f (x) = x 5 - x

f (x) = cos (3 x) - 2, (0, 2 π)

f (x) = \frac{1}{x + 1}

f (x) = - 3 sin (2 x) - 4

f (x, y) = x^{3} - 4 x y + y^{3}

What is the extreme f(x,y)= 1/x-(64)/y+xy ?
The extreme f(x,y)= 1/x-(64)/y+xy is Maximum(-1/4 ,16)