What is the extreme f(x)=e^{4x^2+2y^2-32x} ?

The extreme f(x)=e^{4x^2+2y^2-32x} is Minimum(4,0)

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extreme f(x)=e^{4x^2+2y^2-32x}

extreme points

f (x) = e^{4 x^{2} + 2 y^{2} - 32 x}

Minimum (4, 0)

Solution steps

Find the critical points:

(4, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 (4 e^{4 x^{2} + 2 y^{2} - 32 x} y^{2} + e^{4 x^{2} + 2 y^{2} - 32 x})

D (x, y) = 32 e^{2 (2 x^{2} + y^{2} - 16 x)} (8 x^{2} - 64 x + 129) (4 e^{4 x^{2} + 2 y^{2} - 32 x} y^{2} + e^{4 x^{2} + 2 y^{2} - 32 x}) - 16 e^{2 (4 x^{2} + 2 y^{2} - 32 x)} y^{2} (8 x - 32)^{2}

Check the sign of at each critical point

Check critical point

(4, 0) :

Minimum

Minimum (4, 0)

f (x, y) = x y (8 + x) (y - 9)

\frac{3}{2} x^{2} + x y + \frac{1}{108} y^{4}

(3 x - 4)^{2}

\frac{x ^{3}}{x ^{2} + 4}

f (x) = 2 x^{3} + 3 x^{2} - 36 x - 20

What is the extreme f(x)=e^{4x^2+2y^2-32x} ?
The extreme f(x)=e^{4x^2+2y^2-32x} is Minimum(4,0)