5_3_58.mws

Application of integration:  5.3.58

Given the rate of change in sales

diff(s(t),t) = 2.38*t+10.8/t

where t = years after 1990,

and given that sales in 1992 were 55484,

a) find s(t).

b) find sales in 1998.

>    restart;

>    s1:=t->2.38*t+10.8/t;  #given rate of change in sales at time t

s1 := proc (t) options operator, arrow; 2.38*t+10.8/t end proc

>    Int(s1(t),t);
value(%);  # the symbol % refers to the previous output

Int(2.38*t+10.8/t,t)

1.190000000*t^2+10.80000000*ln(t)

>    s:=%+C;  #name the sales function obtained by integration, with an unknown added constant C
s:=unapply(%,t);

s := 1.190000000*t^2+10.80000000*ln(t)+C

s := proc (t) options operator, arrow; 1.190000000*t^2+10.80000000*ln(t)+C end proc

>    solve(s(2)=55484,C);  #plug in the given FACT to find C

55471.75401

>    C:=%:  #assign C to the previous output

>    s(t);   #our final answer to (a)

1.190000000*t^2+10.80000000*ln(t)+55471.75401

>    evalf(s(2)); #check FACT that was given

55484.00000

>    D(s);  #check derivative

proc (t) options operator, arrow; 2.380000000*t+10.80000000/t end proc

>    s(8);  #evaluate the model at  t = 8 (1998)
evalf(%);   #our answer to (b)

55547.91401+10.80000000*ln(8)

55570.37198