What is the extreme f(x)=4xy+50y-9y^2-1/10 x^2y-130 ?

The extreme f(x)=4xy+50y-9y^2-1/10 x^2y-130 is Maximum(20,5)

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extreme f(x)=4xy+50y-9y^2-1/10 x^2y-130

extreme points

f (x) = 4 x y + 50 y - 9 y^{2} - \frac{1}{10} x^{2} y - 130

Maximum (20, 5)

Solution steps

Find the critical points:

(20, 5)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 18

D (x, y) = \frac{18 y + 8 x}{5} - 16 - \frac{x ^{2}}{25}

Check the sign of at each critical point

Check critical point

(20, 5) :

Maximum

Maximum (20, 5)

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

f (x) = 4 x^{3} - 108 x^{2} + 729 x

f (x) = 4 x^{\frac{3}{4}} - x, 0 \leq x \leq 256

- x^{3} + 6 x^{2} + 15 x + 4

f (x) = x^{2} - 5 x - 1

What is the extreme f(x)=4xy+50y-9y^2-1/10 x^2y-130 ?
The extreme f(x)=4xy+50y-9y^2-1/10 x^2y-130 is Maximum(20,5)