What is the extreme xy^2+8/(xy)-ln(y) ?

The extreme xy^2+8/(xy)-ln(y) is Minimum(64, 1/8)

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extreme xy^2+8/(xy)-ln(y)

extreme points

x y^{2} + \frac{8}{x y} - ln (y)

Minimum (64, \frac{1}{8})

Solution steps

Find the critical points:

(64, \frac{1}{8})

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 x + \frac{16}{x y ^{3}} + \frac{1}{y ^{2}}

D (x, y) = \frac{- 4 x ^{4} y ^{6} + 16 x y + 192}{x ^{4} y ^{4}}

Check the sign of at each critical point

Check critical point

(64, \frac{1}{8}) :

Minimum

Minimum (64, \frac{1}{8})

f (x) = 6 x - 6, - 6 \leq x \leq 3

S (v, d) = 4 π v d + 4 π v^{2}

f (x) = x + \frac{23}{x}

f (x, y) = 7 x^{2} - 9 y^{2} + 2 x + 9 y + 6

U (x, t) = e^{(x^{2} - t^{2})}

What is the extreme xy^2+8/(xy)-ln(y) ?
The extreme xy^2+8/(xy)-ln(y) is Minimum(64, 1/8)