What is the extreme e^{4x^2+2y^2-24x} ?

The extreme e^{4x^2+2y^2-24x} is Minimum(3,0)

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extreme e^{4x^2+2y^2-24x}

extreme points

e^{4 x^{2} + 2 y^{2} - 24 x}

Minimum (3, 0)

Solution steps

Find the critical points:

(3, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 (4 e^{4 x^{2} + 2 y^{2} - 24 x} y^{2} + e^{4 x^{2} + 2 y^{2} - 24 x})

D (x, y) = 32 e^{2 (2 x^{2} + y^{2} - 12 x)} (8 x^{2} - 48 x + 73) (4 e^{4 x^{2} + 2 y^{2} - 24 x} y^{2} + e^{4 x^{2} + 2 y^{2} - 24 x}) - 16 e^{2 (4 x^{2} + 2 y^{2} - 24 x)} y^{2} (8 x - 24)^{2}

Check the sign of at each critical point

Check critical point

(3, 0) :

Minimum

Minimum (3, 0)

f (x) = x^{3} - 3 x^{2} + 9 x + 1

f (x) = - \frac{1}{3} x^{3} + 2 x^{2} + 12 x + 28

f (x) = - 0.015 x^{2} + 1.22 x - 7.2

f (x) = 8 - x

f (x) = 6 + 81 x - 3 x^{3}, 0 \leq x \leq 4

What is the extreme e^{4x^2+2y^2-24x} ?
The extreme e^{4x^2+2y^2-24x} is Minimum(3,0)