Recall that the monthly revenue R in thousands of dollars of

Recall that the monthly revenue R (in thousands of dollars) of a bakery store t months can be modelled by R(t) = 1/12 t^3 - t^2 + 15/4 t, 0 lessthanorequalto t lessthanorequalto 8. We shall find the intervals on which R is increasing or R is decreasing by following the following steps. Find the derivative R\'(t). Find values (if any) of t when R\'(t) = 0. Find values (if any) of t when R\'(t) is undefined. Use the values found in (b) and (c) to test the sign of R\'(t) on some particular intervals.

Solution

We have R(t) = t3/12 –t2 + 15t/4 so that R’(t) = (1/12)* 3t2 -2t +15/4= t2/4 -2t +15/4. If R’(t) = 0, then t2/4 -2t +15/4 =0. On multiplying both the sides by 4, we have t2 -8t +15 = 0 or, t2 -3t -5t+15 = 0 or, t(t-3) -5(t -3) = 0 or, (t-3)(t-5) = 0. Thus , if R’(t) = 0, then t = 3 or, t= 5. Since R’(t) is a polynomial of 2nd degree, it is a continuous function. There is no discontinuiy and R’(t) is defined for all real values of t. If 3 < t < 5, then R’(t) is negative( R’(3.5) = -0.1875, R’(4) = -0.25 etc.). If t < 3, then R’(t) is positive( R’(2) = 0.75, R’(0) = 3.75, R’(-1) = 6 etc.). If t > 5, then R’(t) is positive( R’(6)= 0.75, R’(7) = 2 etc.). Thus, R’(t) > 0 if t (- ,3) , R’(t) < 0 if t (3,5) and R’(t) > 0 if t (5,).