What is the extreme U(x,y)=-x^2+8-y^2 ?

The extreme U(x,y)=-x^2+8-y^2 is Maximum(0,0)

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extreme U(x,y)=-x^2+8-y^2

extreme points

U (x, y) = - x^{2} + 8 - y^{2}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 2 x^{4} + 8 x^{2} - 8 x^{3} - 1

f (x) = 9 x^{3} - 54 x^{2} + 81 x + 13, (- 6, 2)

f (x) = x^{3} - 12 x^{2} + 45 x + 7

f (x) = x^{7} (x + 4)^{4}, - 10 \leq x \leq 13

y = x^{4} - 8 x^{2} + 5

What is the extreme U(x,y)=-x^2+8-y^2 ?
The extreme U(x,y)=-x^2+8-y^2 is Maximum(0,0)