19.mws

>    restart; problem #19

>    N:=t->-.08*t^3+.4*t^2+8.08*t;  #define our function

N := proc (t) options operator, arrow; -.8e-1*t^3+.4*t^2+8.08*t end proc

>    N1:=D(N); #find derivative

N1 := proc (t) options operator, arrow; -.24*t^2+.8*t+8.08 end proc

>    N2:=D(N1);  #find second deriv

N2 := proc (t) options operator, arrow; -.48*t+.8 end proc

>    solve(N2(t)=0,t);  #find inflection points

1.666666667

>    T:=%; #our answer, in time after 3:00

T := 1.666666667

>    (T-1)*60;  #convert decimal part of hour to minutes

40.0000000

Conclusion: 3:00 + 1:40 = 4:40 = time of maximal rate

-----------visual verification:

>    plot(N1(t),t=0..8,title="rate of assembly");

[Maple Plot]