What is the critical f(x,y)=xy+ln(x)+32y^2 ?

The critical f(x,y)=xy+ln(x)+32y^2 is (-8, 1/8),(8,-1/8)

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critical f(x,y)=xy+ln(x)+32y^2

critical points

f (x, y) = x y + ln (x) + 32 y^{2}

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

Solution steps

x y + ln (x) + 32 y^{2}

Find where

\nabla f (x, y)

is equal to zero or undefined

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

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

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

f (x) = 5 x^{3} - 3 x^{5}

\frac{x}{x ^{2} + 11 x + 30}

f (x) = (8 - 6 x) e^{x}

What is the critical f(x,y)=xy+ln(x)+32y^2 ?
The critical f(x,y)=xy+ln(x)+32y^2 is (-8, 1/8),(8,-1/8)