What is the extreme f(x,y)=e^{(x^2+0.5y^2-5xy-3x)} ?

The extreme f(x,y)=e^{(x^2+0.5y^2-5xy-3x)} is Saddle(-3/23 ,-15/23)

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extreme f(x,y)=e^{(x^2+0.5y^2-5xy-3x)}

extreme points

f (x, y) = e^{(x^{2} + 0.5 y^{2} - 5 x y - 3 x)}

Saddle (- \frac{3}{23}, - \frac{15}{23})

Solution steps

Find the critical points:

(- \frac{3}{23}, - \frac{15}{23})

\frac{\partial ^{2} f}{\partial x ^{2}} = e^{x^{2} - 5 x y + 0.5 y^{2} - 3 x} ((y - 5 x)^{2} + 1)

D (x, y) = e^{2 x^{2} - 10 x y - 6 x + y^{2}} (4 x^{2} - 20 x y - 12 x + 25 y^{2} + 30 y + 11) ((y - 5 x)^{2} + 1) - e^{2 (x^{2} - 5 x y + 0.5 y^{2} - 3 x)} ((y - 5 x) (- 3 + 2 x - 5 y) - 5)^{2}

Check the sign of at each critical point

Check critical point

(- \frac{3}{23}, - \frac{15}{23}) :

Saddle

Saddle (- \frac{3}{23}, - \frac{15}{23})

f (x) = x^{2} - 15 x + 56

y = x^{4} - x^{3}

f (t) = t u (t - 3)

f (x) = 6 x - 12

2 x^{5} + 5 x^{4}, - 2 \leq x \leq 1

What is the extreme f(x,y)=e^{(x^2+0.5y^2-5xy-3x)} ?
The extreme f(x,y)=e^{(x^2+0.5y^2-5xy-3x)} is Saddle(-3/23 ,-15/23)