What is the extreme f(x,y)=16x+12y-2x^2-3y^2 ?

The extreme f(x,y)=16x+12y-2x^2-3y^2 is Maximum(4,2)

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extreme f(x,y)=16x+12y-2x^2-3y^2

extreme points

f (x, y) = 16 x + 12 y - 2 x^{2} - 3 y^{2}

Maximum (4, 2)

Solution steps

Find the critical points:

(4, 2)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(4, 2) :

Maximum

Maximum (4, 2)

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

x^{3} - 12 x + 3

f (x, y) = x y + \frac{3}{x} + \frac{9}{y}

\frac{4 x}{x ^{2} - 9}

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

What is the extreme f(x,y)=16x+12y-2x^2-3y^2 ?
The extreme f(x,y)=16x+12y-2x^2-3y^2 is Maximum(4,2)