What is the extreme f(24/13 , 88/39)=3xy-4x^2-3y^2+8x+8y-4 ?

The extreme f(24/13 , 88/39)=3xy-4x^2-3y^2+8x+8y-4 is Maximum(24/13 , 88/39)

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extreme f(24/13 , 88/39)=3xy-4x^2-3y^2+8x+8y-4

extreme points

f (\frac{24}{13}, \frac{88}{39}) = 3 x y - 4 x^{2} - 3 y^{2} + 8 x + 8 y - 4

Maximum (\frac{24}{13}, \frac{88}{39})

Solution steps

Find the critical points:

(\frac{24}{13}, \frac{88}{39})

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

D (x, y) = 39

Check the sign of at each critical point

Check critical point

(\frac{24}{13}, \frac{88}{39}) :

Maximum

Maximum (\frac{24}{13}, \frac{88}{39})

- x^{3} + 75 x + 80

f (x) = x^{3} + 6 x^{2} + 5

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

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

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

What is the extreme f(24/13 , 88/39)=3xy-4x^2-3y^2+8x+8y-4 ?
The extreme f(24/13 , 88/39)=3xy-4x^2-3y^2+8x+8y-4 is Maximum(24/13 , 88/39)