Optimizing Water Flow: Finding Maximum Mass and Rate of Change

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lgen0290
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Homework Statement



Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 6.00t 0.8 - 3.35t + 23.00, with t 0, m in grams, and t in seconds.
(a) At what time is the water mass greatest?

(b) What is that greatest mass?

(c) In kilograms per minute, what is the rate of mass change at t = 2.00 s?

(d) In kilograms per minute, what is the rate of mass change at t = 5.00 s?

Homework Equations


The first derivative would be 4.8t^-.2-335


The Attempt at a Solution



a)I set the derivative equal to zero and figured t to be 2.3375, but it says that's wrong.
b)I assume I'd plug a into the original and try tht, but I can't get a.
c and d)I tried to put 2 and 5 into t as the original, but they are not right.
 
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a) is, as you said, the solution for t when

[tex]\frac{dm}{dt}=0.[/tex]

However, the solution to

[tex]t^{0.2}=\left[\frac{4.8}{3.35}\right][/tex]

is not 2.3375, check your algebra.

b), as you said is [tex]m(t)[/tex] for the solution above
c) and d) can both be found by substituting the times into the expression for [tex]\frac{dm}{dt}.[/tex]
 
Thanks. Where did the 3.35 come from? How would I solve t^-.2?
 
The 3.35 is from the original expression for m, unless you've mistyped it. I used logarithms and the change of base rule to solve for t.