What is the extreme f(x,y)=60x-2x^2-3y^2+30y ?

The extreme f(x,y)=60x-2x^2-3y^2+30y is Maximum(15,5)

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extreme f(x,y)=60x-2x^2-3y^2+30y

extreme points

f (x, y) = 60 x - 2 x^{2} - 3 y^{2} + 30 y

Maximum (15, 5)

Solution steps

Find the critical points:

(15, 5)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(15, 5) :

Maximum

Maximum (15, 5)

\frac{x ^{2}}{x - 4}

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

f (x, y) = ∣ x - y ∣

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

f (x) = 10 x^{2} + 36 x + 36

What is the extreme f(x,y)=60x-2x^2-3y^2+30y ?
The extreme f(x,y)=60x-2x^2-3y^2+30y is Maximum(15,5)