Asked by charlotte
                Let t be time in seconds and let r(t) be the rate, in gallons per second, that water enters a reservoir:
r(t)=700−40t.
a) For 0≤t≤30, when does the reservoir have the most water?
b) For 0≤t≤30, when does the reservoir have the least water?
            
        r(t)=700−40t.
a) For 0≤t≤30, when does the reservoir have the most water?
b) For 0≤t≤30, when does the reservoir have the least water?
Answers
                    Answered by
            Steve
            
    if a(t) is the amount of water at time t, and the tank is initially empty,
da/dt = r(t) = 700-40t
a(t) = 700t-20t^2
a(t) is a parabola with vertex at t=700/40=17.5
So, that's the time of max water.
For minimum, find whether a(30) is positive or negative. Recall that a(0) = 0.
    
da/dt = r(t) = 700-40t
a(t) = 700t-20t^2
a(t) is a parabola with vertex at t=700/40=17.5
So, that's the time of max water.
For minimum, find whether a(30) is positive or negative. Recall that a(0) = 0.
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