What is the extreme f(x)=4y^2-5x=2y-3y^2 ?

The extreme f(x)=4y^2-5x=2y-3y^2 is Maximum(1/3 , 1/3)

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extreme f(x)=4y^2-5x=2y-3y^2

extreme

Maximum (\frac{1}{3}, \frac{1}{3})

Solution steps

f^{'} (y) = 2 - 6 y

Find intervals:

Increasing

: - \infty < y < \frac{1}{3},

Decreasing

: \frac{1}{3} < y < \infty

Plug

y = \frac{1}{3}

into

2 y - 3 y^{2} : \frac{1}{3}

Maximum (\frac{1}{3}, \frac{1}{3})

What is the extreme f(x)=4y^2-5x=2y-3y^2 ?
The extreme f(x)=4y^2-5x=2y-3y^2 is Maximum(1/3 , 1/3)