What is the extreme f(x,y)=ln(2x+y)-x^2-4y ?

The extreme f(x,y)=ln(2x+y)-x^2-4y is Maximum(4,-31/4)

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extreme f(x,y)=ln(2x+y)-x^2-4y

extreme points

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

Maximum (4, - \frac{31}{4})

Solution steps

Find the critical points:

(4, - \frac{31}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{1}{( 2 x + y ) ^{2}}

D (x, y) = \frac{2}{( y + 2 x ) ^{2}}

Check the sign of at each critical point

Check critical point

(4, - \frac{31}{4}) :

Maximum

Maximum (4, - \frac{31}{4})

U (X, Y) = X^{0.4} Y^{0.6}

f (x) = \frac{2 x + 4}{x + 1}

pp + qq

f (x, y) = 18 x y + 2500

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

What is the extreme f(x,y)=ln(2x+y)-x^2-4y ?
The extreme f(x,y)=ln(2x+y)-x^2-4y is Maximum(4,-31/4)