Given
... y = -0.5|t -288| +144
... t is minutes after 10 am
Find
... a) t such that y = 100
... b) clock time corresponding to those values of t
... c) t such that y is a maximum
... d) clock time corresponding to that maximum
Solution
a) y = 100
... -0.5|t -288| +144 = 100 . . . . . substitute for y
... -0.5|t -288| = -44 . . . . . . . . . . subtract 144
... |t - 288| = 88 . . . . . . . . . . . . . multiply by -2
... -88 = t -288 . . . . . . . one meaning of the above equation
... t = 200 . . . . . . . . . . . add 288
... t -288 = 88 . . . . . . . . the other meaning of the above equation
... t = 376 . . . . . . . . . . . add 288
There are 100 shoppers in the store at t=200 and t=376
b) On a 24-hour clock, the clock time (h) corresponding to a value of t is ...
... h(t) = 10 + t/60
... h(200) = 10 + 200/60 = 13 1/3 . . . . corresponds to 1:20 pm
... h(376) = 10 + 376/60 = 16 4/15 . . . . corresponds to 4:16 pm
c) The value of y is a maximum where the content of the absolute value is 0. Thus the maximum number of shoppers is 144.
d) The maximum number of shoppers occurs when t=288, so at clock time ...
... h(288) = 10 + 288/60 = 14 4/5 . . . . corresponds to 2:48 pm
_____
An hour is 60 minutes, so 1/3 hour is ...
... 1/3×60 = 20 minutes; and ...
... 4/15×60 = 16 minutes;
... 4/5×60 = 48 minutes.
