What is the extreme f(x,y)=60x+34y-4xy-6x^2-3y^2+5 ?

The extreme f(x,y)=60x+34y-4xy-6x^2-3y^2+5 is Maximum(4,3)

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extreme f(x,y)=60x+34y-4xy-6x^2-3y^2+5

extreme points

f (x, y) = 60 x + 34 y - 4 x y - 6 x^{2} - 3 y^{2} + 5

Maximum (4, 3)

Solution steps

Find the critical points:

(4, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = 56

Check the sign of at each critical point

Check critical point

(4, 3) :

Maximum

Maximum (4, 3)

f (x) = x^{2} \cdot e^{- x^{2}}

f (x, y) = x^{4} + \frac{y ^{4}}{x} y

2 + 3 x + 27 x^{- 1}

f (x) = x^{3} - 12 x^{2} + 45 x + 2

f (x) = - 5 x^{3} - 4 x^{4}

What is the extreme f(x,y)=60x+34y-4xy-6x^2-3y^2+5 ?
The extreme f(x,y)=60x+34y-4xy-6x^2-3y^2+5 is Maximum(4,3)