What is the extreme f(x,y)=(x+2y-7)^2+(2x+y-5)^2 ?

The extreme f(x,y)=(x+2y-7)^2+(2x+y-5)^2 is Minimum(1,3)

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extreme f(x,y)=(x+2y-7)^2+(2x+y-5)^2

extreme points

f (x, y) = (x + 2 y - 7)^{2} + (2 x + y - 5)^{2}

Minimum (1, 3)

Solution steps

Find the critical points:

(1, 3)

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

D (x, y) = 36

Check the sign of at each critical point

Check critical point

(1, 3) :

Minimum

Minimum (1, 3)

f (x) = x^{2} + 9 x - 8

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

f (x) = x^{2} + \frac{520}{x}

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

2 x^{3} - 2 x^{2} - 2 x + 3

What is the extreme f(x,y)=(x+2y-7)^2+(2x+y-5)^2 ?
The extreme f(x,y)=(x+2y-7)^2+(2x+y-5)^2 is Minimum(1,3)