What is the extreme f(x,y)=120x+120y-xy-x^2-y^2 ?

The extreme f(x,y)=120x+120y-xy-x^2-y^2 is Maximum(40,40)

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extreme f(x,y)=120x+120y-xy-x^2-y^2

extreme points

f (x, y) = 120 x + 120 y - x y - x^{2} - y^{2}

Maximum (40, 40)

Solution steps

Find the critical points:

(40, 40)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(40, 40) :

Maximum

Maximum (40, 40)

f (x) = (x + 7)^{\frac{2}{3}}

f (x) = - x + 2 cos (x)

\frac{x + 4}{x ^{2} - 5 x - 36}

f (x) = - \frac{5 x ^{3}}{3} + 10 x^{2} - 15 x, (- 2, 5)

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

What is the extreme f(x,y)=120x+120y-xy-x^2-y^2 ?
The extreme f(x,y)=120x+120y-xy-x^2-y^2 is Maximum(40,40)