What is the extreme f(x,y)= 1/x+xy-8/y ?

The extreme f(x,y)= 1/x+xy-8/y is Maximum(-1/2 ,4)

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extreme f(x,y)= 1/x+xy-8/y

extreme points

f (x, y) = \frac{1}{x} + x y - \frac{8}{y}

Maximum (- \frac{1}{2}, 4)

Solution steps

Find the critical points:

(- \frac{1}{2}, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{16}{y ^{3}}

D (x, y) = - \frac{32}{x ^{3} y ^{3}} - 1

Check the sign of at each critical point

Check critical point

(- \frac{1}{2}, 4) :

Maximum

Maximum (- \frac{1}{2}, 4)

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

y = 3 x^{2} - 12 x + 13

f (x) = \frac{1}{4} x^{4} - 2 x^{3} + 3

f (x) = \frac{( - 8 x ^{3} + 3 x ^{2} )}{4}

f (x) = 2 x^{3} + x^{2} - 4 x + 4

What is the extreme f(x,y)= 1/x+xy-8/y ?
The extreme f(x,y)= 1/x+xy-8/y is Maximum(-1/2 ,4)