What is the extreme f(p,q)=p^2-2p+0.8q^2 ?

The extreme f(p,q)=p^2-2p+0.8q^2 is Minimum(1,0)

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extreme f(p,q)=p^2-2p+0.8q^2

extreme points

f (p, q) = p^{2} - 2 p + 0.8 q^{2}

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial p ^{2}} = 1.6

D (p, q) = 3.2

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

y = 2 + 12 x - x^{3}

y = \frac{x ^{3}}{3} - 2 x^{2} - 5 x

f (x) = x^{3} - 3 x^{2} + 12, - 2 \leq x \leq 4

f (x) = \frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x

f (x) = 6 x + 12 y - x^{2} - 3 y^{2}

What is the extreme f(p,q)=p^2-2p+0.8q^2 ?
The extreme f(p,q)=p^2-2p+0.8q^2 is Minimum(1,0)