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

The extreme f(x,y)=e^{2x-4y-x^2-y^2} is Maximum(1,-2)

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

extreme points

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

Maximum (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

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

D (x, y) = 4 e^{- 2 x^{2} + 4 x - 2 y^{2} - 8 y} (2 x^{2} - 4 x + 1) (2 y^{2} + 8 y + 7) - e^{2 (2 x - 4 y - x^{2} - y^{2})} (- 2 x + 2)^{2} (- 2 y - 4)^{2}

Check the sign of at each critical point

Check critical point

(1, - 2) :

Maximum

Maximum (1, - 2)

f (x) = - 4 x^{2} + 172 x - 1528

f (x) = x (5 - x) (8 - x)

f (x) = x^{4} - 8 x^{3} + 16 x^{2} + 6

f (x) = x^{4} - 8 x^{3} + 16 x^{2} + 7

f (x) = 2 x^{2} - 10 x + 5, 0 \leq x \leq 7

What is the extreme f(x,y)=e^{2x-4y-x^2-y^2} ?
The extreme f(x,y)=e^{2x-4y-x^2-y^2} is Maximum(1,-2)