What is the extreme f(x,y)=kxy^2+kx^2y+xy ?

The extreme f(x,y)=kxy^2+kx^2y+xy is Saddle(0,-1/k),Maximum(-1/(3k),-1/(3k))

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extreme f(x,y)=kxy^2+kx^2y+xy

extreme points

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

Saddle (0, - \frac{1}{k}), Maximum (- \frac{1}{3 k}, - \frac{1}{3 k})

Solution steps

Find the critical points:

(0, - \frac{1}{k}), (- \frac{1}{3 k}, - \frac{1}{3 k})

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 k x

D (x, y) = - 4 k^{2} x^{2} - 4 k x - 4 k^{2} x y - 4 k^{2} y^{2} - 4 k y - 1

Check the sign of at each critical point

Check critical point

(0, - \frac{1}{k}) :

Saddle

Check critical point

(- \frac{1}{3 k}, - \frac{1}{3 k}) :

Maximum

Saddle (0, - \frac{1}{k}), Maximum (- \frac{1}{3 k}, - \frac{1}{3 k})

y = 2 x^{3} + 3 x^{2} - 12 x + 5

f (x) = 4 x + 4 sin (x), 0 \leq x \leq 2 π

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

f (x) = 2 x^{2} - 16 ln (x)

f (x, y) = x 1 + y^{3}

What is the extreme f(x,y)=kxy^2+kx^2y+xy ?
The extreme f(x,y)=kxy^2+kx^2y+xy is Saddle(0,-1/k),Maximum(-1/(3k),-1/(3k))