What is the extreme f(x,y)=(x^3)/3-y^2-5x+xy-3 ?

The extreme f(x,y)=(x^3)/3-y^2-5x+xy-3 is Saddle(2,1),Maximum(-5/2 ,-5/4)

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extreme f(x,y)=(x^3)/3-y^2-5x+xy-3

extreme points

f (x, y) = \frac{x ^{3}}{3} - y^{2} - 5 x + x y - 3

Saddle (2, 1), Maximum (- \frac{5}{2}, - \frac{5}{4})

Solution steps

Find the critical points:

(2, 1), (- \frac{5}{2}, - \frac{5}{4})

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

D (x, y) = - 4 x - 1

Check the sign of at each critical point

Check critical point

(2, 1) :

Saddle

Check critical point

(- \frac{5}{2}, - \frac{5}{4}) :

Maximum

Saddle (2, 1), Maximum (- \frac{5}{2}, - \frac{5}{4})

h (x) = - 2 x^{2} - 1, - 4 \leq x \leq 3

x^{3} - 12 x^{2} + 45 x - 4

y^{4} - y^{2} - 2 x y + x^{2}

f (x, y) = lo g_{10} (\frac{( x ^{2} + 2 \cdot y )}{( x + y ) ^{2}})

f (x) = y = tan (\frac{π x}{8}), 0 \leq x \leq 2

What is the extreme f(x,y)=(x^3)/3-y^2-5x+xy-3 ?
The extreme f(x,y)=(x^3)/3-y^2-5x+xy-3 is Saddle(2,1),Maximum(-5/2 ,-5/4)