What is the extreme f(x,y)=e^{4y-y^2-x^2} ?

The extreme f(x,y)=e^{4y-y^2-x^2} is Maximum(0,2)

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extreme f(x,y)=e^{4y-y^2-x^2}

extreme points

f (x, y) = e^{4 y - y^{2} - x^{2}}

Maximum (0, 2)

Solution steps

Find the critical points:

(0, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{4 y - x^{2} - y^{2}} (2 y^{2} - 8 y + 7)

D (x, y) = - 4 e^{4 y - x^{2} - y^{2}} (- 2 e^{4 y - y^{2} - x^{2}} x^{2} + e^{4 y - y^{2} - x^{2}}) (2 y^{2} - 8 y + 7) - 4 e^{2 (4 y - y^{2} - x^{2})} x^{2} (- 2 y + 4)^{2}

Check the sign of at each critical point

Check critical point

(0, 2) :

Maximum

Maximum (0, 2)

f (x) = sin (x) + sin^{3} (x), - π < x < π

y = (x - 1)^{3} (x - 5)

f (x) = 4 x - 6

f (x) = - \frac{2}{x ^{2}}, 0.5 \leq x \leq 2

f (x) = - \frac{2}{x ^{2}}, 0.5 \leq x \leq 5

What is the extreme f(x,y)=e^{4y-y^2-x^2} ?
The extreme f(x,y)=e^{4y-y^2-x^2} is Maximum(0,2)