What is the extreme P(x,y)=0.3+x^2+0.2y^2+0.1xy-14x-10y+2000 ?

The extreme P(x,y)=0.3+x^2+0.2y^2+0.1xy-14x-10y+2000 is Minimum(5.82278…, 1860/79)

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extreme P(x,y)=0.3+x^2+0.2y^2+0.1xy-14x-10y+2000

extreme points

P (x, y) = 0.3 + x^{2} + 0.2 y^{2} + 0.1 x y - 14 x - 10 y + 2000

Minimum (5.82278 \dots, \frac{1860}{79})

Solution steps

Find the critical points:

(5.82278 \dots, \frac{1860}{79})

\frac{\partial ^{2} f}{\partial x ^{2}} = 0.4

D (x, y) = 0.79

Check the sign of at each critical point

Check critical point

(5.82278 \dots, \frac{1860}{79}) :

Minimum

Minimum (5.82278 \dots, \frac{1860}{79})

f (x, y) = \frac{1}{2} x^{2} + \frac{1}{2} y^{2} - x^{2} y + 5

y = \frac{x ^{4}}{12} - \frac{x ^{3}}{2} + x^{2} + 10

f (x) = (x - 2)^{7} (x + 4)^{7}

sin^{8} (x) + cos^{8} (x)

f (x, y) = 400 - 81 x^{2} - 4 y^{2}

What is the extreme P(x,y)=0.3+x^2+0.2y^2+0.1xy-14x-10y+2000 ?
The extreme P(x,y)=0.3+x^2+0.2y^2+0.1xy-14x-10y+2000 is Minimum(5.82278…, 1860/79)