Posted by Jan on Sunday, October 13, 2013 at 11:33pm.
Find the critical numbers of the function.
h(t)=t^3/46^1/4

Calculus  Typo  Steve, Monday, October 14, 2013 at 12:02am
I think there's a typo there. Anyway, just find where h'=0.

Calculus  Jan, Monday, October 14, 2013 at 1:23am
oh, sorry about that!
h(t)=t^3/46t^1/4

Calculus  Steve, Monday, October 14, 2013 at 4:12am
well, dh/dt = 3t^2(11∜t8) / 8(23∜t)^2
We want dh/dt=0, so t=0 or t=(8/11)^4
h(t) and dh/dt are undefined where t=(2/3)^4
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