Question

p(t) hundred dollars per year p(t) = 0.6677-121+ 40t + 80 gives the rate of change of profit in hundred dollars per year) at Java Joe's, t years after 2000,0 St<12. Check: p(4)= 90.688 Use this information to answer the next six questions. 8 9 10 11 12 1. Between 2000 and 2012, the profit was highest in [Select] • hundred dollars. II. From 2005 to 2010, the profit at Java Joe's decreased by 36.719 III. The graph of the accumulation function Acx) = 5 *ptt)dt, is decreasing at a slower rate on the interval

Answer 1

solution: → Given, 3 plt): 0.667 -12++ 404 +80 gives rate of change t years aften Oct212 2000 ☺ Tare, pic+) = 26-24t +40 2+12+ +20) = 0 (t-10) (+-2) = 0 points t=2, ç pll (t): 46-24 p''(t) = 4(2) – 24 talo critical = 8-24 at t: 2 2000+2= 2002 Here, 10 so, profit and higher in s (0.667 +3-1977 -12++ tot +80) dt 0.667 +4- ++++ 20++ 80+], - (0.467 cccup20. v'tte (1) -{ "-667 (5)£ 46579420(5)80(5) 467.5 • 504.219 -36.719 36. 719 hundred dollung sol cer) profit decreased by Take A(4): Sip() dt Aca)= P(a) .

A(a)= = P(x) = 0.667 42-126 + 40t +80=0 for critical point 50-667 x3-12x3+ 40x+80=0 caladetor (47512) x= 7.022 + 12 7.022 A6) decreasing since 7-002 iş critical the interve 10.7022). point, so Ac) decreasing Aca) slow er 2 cxc6 at slow rate hean 7022 So at rate a decreasing Pu) Take, AC = 2*3-247 +40 = 0, 2312*+20)=0 x=2,10 now sign check -ve tve +ve 2 10 14 Aco) Ah (1) 40 A'(3) = 18 - 72+40= = 242-264+40 = 18 so, between 2 and sign is negative Hence cancque down 2<x< 10 10 Take A(x)= S p(t) at s(0.66723_1277 400+80) at (::A(0)=500)=> A(x)= 0-667 +4 ► A(O) 4++ 200+800 0.667 (0)4 - 46032+2060+80(2)+c 4

500 0-0 +0+0+0 (=500 0.667 and, → A(X) = +4- 44+ 20t'7 80+ +500 0.664 ()- 462+2002) 780(2)+500 AC2). : 710.7 So, answer is $710.7 dollay (NP), Here, signed area 0-667 +4 2+206780t 4 = 386 Total $. area = + 443 4 !* 59:- 157 Spo ſo.66774 ht'+20++80+] 710,667 ** +20t180t 10.667 (+3€ AC+F+20 (4578067] 110-664 (12)440125420012578 06127) 12 n 7 7 o (0.667 4 568.366+ ll 386-568.366|| = 568-336+ 18 2.366 750.732 775) so, Total signed area between pt) and t-aris ra 386 1 and total area between P(t) and t-araş iş 751