What is the extreme f(4)=e^{2x^2+4y^2-16x} ?

The extreme f(4)=e^{2x^2+4y^2-16x} is Minimum(4,0)

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extreme f(4)=e^{2x^2+4y^2-16x}

extreme points

f (4) = e^{2 x^{2} + 4 y^{2} - 16 x}

Minimum (4, 0)

Solution steps

Find the critical points:

(4, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 (8 e^{2 x^{2} + 4 y^{2} - 16 x} y^{2} + e^{2 x^{2} + 4 y^{2} - 16 x})

D (x, y) = 32 e^{2 (x^{2} + 2 y^{2} - 8 x)} (4 x^{2} - 32 x + 65) (8 e^{2 x^{2} + 4 y^{2} - 16 x} y^{2} + e^{2 x^{2} + 4 y^{2} - 16 x}) - 64 e^{2 (2 x^{2} + 4 y^{2} - 16 x)} y^{2} (4 x - 16)^{2}

Check the sign of at each critical point

Check critical point

(4, 0) :

Minimum

Minimum (4, 0)

(4 x + y - 2)^{2} + (x + y - 1)^{2} + (4 - 4 x - y)^{2}

y = 5 cos^{2} (x)

h (x, t) = 1.5 + lo g_{3} (x) (t + 1)

\frac{3200}{x} + 200 + 50 x

f (x y) = x^{2} + 4 y^{2}

What is the extreme f(4)=e^{2x^2+4y^2-16x} ?
The extreme f(4)=e^{2x^2+4y^2-16x} is Minimum(4,0)