## Answers

I am assuming you need solution of part C only as part A and B are correctly solved!

c(t) = 5t 1 27+t3 In order to find when the concentration, we will rese concept q maxima and minima of calcules. first derivative Taking of cct) c'(t) = a cast) o deve

3 d'(t) = 5 cm (2876) ckey = 5(221 at the (1274837 end (27+372 3 c'(t) = 5[(27+43) (1) -+10+3t?] (274832

^c't) = 5/27+63 -3 +] (27+€32 56 (t) = 5 (27 - 2+3) 127+ + 2 for maximum or minimum cl (t) = 0 = 5(27-2€ o ( 27+32 = 27-2ť = 0 263 - 27 3 + = (27.)"

t = 3 (2)/ c(t) will be maximum at to Byte if c" (és va co So, taking second derivative of c(t) c"(t) = d c'(t) dt

-dr5(27-2+3) 7 Duoef ks 24? (29

sc'tt) = 57(2346) (27-24) 1 -(27-28) d(27++327 - ((27+ +32 2

zlity = 5 (27te) (6-2 (3) - 1 l (27-2)(2+)). (27437 (27 +23) 34

5"(t) = 5 1-6++(27+esj?= 2(27-28).

(27+€3).(0+3+²) 7 (27++3;4 at t=3 (2) 1/3

| (중) - - (2) - 227-19 1태를녀름에 223 (24t

N LG x 9. 3. 14 | 2f 22 el) - FI-12 ')-12-3 2. ' (2 )()7 (27끌다 (6) - 1 2 ) - 2(2747)(27)() (27 )

(0) 일) (24 ) ]

re"(一)--卷(!) 7(高) (4) 一 ) 十 ) = "(): 本

-) - - 0. 1036 le cul ) < 0 so (lt) és maximum at t = 3 21/3

.
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