What is the extreme-1/80 t^2+9/40 t-17/80 ?

The extreme-1/80 t^2+9/40 t-17/80 is Maximum(9, 4/5)

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extreme-1/80 t^2+9/40 t-17/80

extreme

Maximum (9, \frac{4}{5})

Solution steps

f^{'} (t) = \frac{- t + 9}{40}

Find intervals:

Increasing

: - \infty < t < 9,

Decreasing

: 9 < t < \infty

Plug

t = 9

into

- \frac{1}{80} t^{2} + \frac{9}{40} t - \frac{17}{80} : \frac{4}{5}

Maximum (9, \frac{4}{5})

What is the extreme-1/80 t^2+9/40 t-17/80 ?
The extreme-1/80 t^2+9/40 t-17/80 is Maximum(9, 4/5)