What is the extreme f(x)=-20x^2-30y^2+560x+360y+180 ?

The extreme f(x)=-20x^2-30y^2+560x+360y+180 is Maximum(14,6)

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extreme f(x)=-20x^2-30y^2+560x+360y+180

extreme

Maximum (14, 6)

Solution steps

Find the critical points:

(14, 6)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 60

D (x, y) = 2400

Check the sign of at each critical point

Check critical point

(14, 6) :

Maximum

Maximum (14, 6)

What is the extreme f(x)=-20x^2-30y^2+560x+360y+180 ?
The extreme f(x)=-20x^2-30y^2+560x+360y+180 is Maximum(14,6)