What is the extreme f(x,y)=x^2+y^2-10x+16y-6 ?

The extreme f(x,y)=x^2+y^2-10x+16y-6 is Minimum(5,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=x^2+y^2-10x+16y-6

extreme points

f (x, y) = x^{2} + y^{2} - 10 x + 16 y - 6

Minimum (5, - 8)

Solution steps

Find the critical points:

(5, - 8)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(5, - 8) :

Minimum

Minimum (5, - 8)

f (x) = e^{- \frac{( x - 3 ) ^{2}}{2}} (- x + 3)

\frac{8 x ^{3}}{3} + 6 x^{2} - 8 x, - 3 \leq x \leq 1

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

f (x) = cos (x) - 9 x, 0 \leq x \leq 4 π

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

What is the extreme f(x,y)=x^2+y^2-10x+16y-6 ?
The extreme f(x,y)=x^2+y^2-10x+16y-6 is Minimum(5,-8)