What is the extreme f(x,y)=ln(4x+y)-x^{(2)}-3y ?

The extreme f(x,y)=ln(4x+y)-x^{(2)}-3y is Maximum(6,-71/3)

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

extreme points

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

Maximum (6, - \frac{71}{3})

Solution steps

Find the critical points:

(6, - \frac{71}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{1}{( 4 x + y ) ^{2}}

D (x, y) = \frac{2}{( y + 4 x ) ^{2}}

Check the sign of at each critical point

Check critical point

(6, - \frac{71}{3}) :

Maximum

Maximum (6, - \frac{71}{3})

\frac{1}{x ^{2} + 2 x}

\frac{e ^{x}}{3 + e ^{x}}

f (x) = 320 x^{2} - 1920 x^{3}

f (x) = \frac{1}{3} x^{3} + 2 x^{2} - 5 x - 4

What is the extreme f(x,y)=ln(4x+y)-x^{(2)}-3y ?
The extreme f(x,y)=ln(4x+y)-x^{(2)}-3y is Maximum(6,-71/3)