What is the extreme f(x,y)=-8-6x+3x^2+8y+5y^2 ?

The extreme f(x,y)=-8-6x+3x^2+8y+5y^2 is Minimum(1,-4/5)

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extreme f(x,y)=-8-6x+3x^2+8y+5y^2

extreme points

f (x, y) = - 8 - 6 x + 3 x^{2} + 8 y + 5 y^{2}

Minimum (1, - \frac{4}{5})

Solution steps

Find the critical points:

(1, - \frac{4}{5})

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 60

Check the sign of at each critical point

Check critical point

(1, - \frac{4}{5}) :

Minimum

Minimum (1, - \frac{4}{5})

f (x) = \frac{4 + x}{1 - x}, - 1 \leq x \leq 0

f (x) = 2 x^{3} - 3 x^{2} - 72 x + 3, - 4 \leq x \leq 5

\frac{x y}{e ^{x^{2} + y^{2}}}

f (x) = 4 y^{3} + x^{2} - 12 y - 36 y + 2

x ln (2 x)

What is the extreme f(x,y)=-8-6x+3x^2+8y+5y^2 ?
The extreme f(x,y)=-8-6x+3x^2+8y+5y^2 is Minimum(1,-4/5)