Question

10&11

10) Find the number of units that must be produced and sold in order to yield the maximum profit, give following equations for revenue and cost: R(x) = 3x C(x) = 0.001x2 +0.7x + 10. A) 1150 units B) 1850 units C) 3700 units D) 2300 units 11) A bookstore has an annual demand for 28,000 copies of a best-selling book. It costs $0.50 to store one copy for one year, and it costs $65 to place an order. Find the optimum number of copies per order. A) 2428 copies B) 2698 copies C) 3257 copies D) 3816 copies

Answer 1

i o profit function 1 = Revenue functia - cost function > PCX) = R(X) – cca > Paa) = 32-[0.001%+0.77 710] 3 P(X) = 32-0100124-0.7x10 => x = 2.3x - 2.00|x−10

To maximise the profit, we will use concept of maxima and minima of calculus. Taking first derivative of P(x) plena com PCM splex)= aha (2-33-0100|x? –10) oplca)=2.3 d (n) = 0.001 a (x²) an dx C10) dx

Рех)= 2+14 () — О-оој ( 2x) - о >Р' си ) = 2+1 — 0.002 и л от тщкі тим ох тиштим : Рcx) = 0 52. 1- ооо2 х го — 2. 2 = ooo 2 20-оо2 х = 22 2 х = 2+3 в ОО 2 > X= (( So

P(n) will be maximum at u= 1150 if pil(n) is - p" (n) < o at x=1150 so, Taking second disivarie Pix) pl(a) = de Plans an puch) & (23 – 010024) 23p"ca) = (2-3) - da 0.002 d (2) dx

MO 3 p"cx) = 0 -0.00201) plex)=-0.002 atx=1150 pl|(1158)= -0.002<0

So, Pon) is maximum alt x = 11 50 So, 1150 units will maximum profits (A) 110 (B) 1850 (②) 3700 (0) 2300 so option (A) is correct!

het a be the number of copies per order. Demand=28000 copies Neember q orders = 28000 x ordering cost = Number of orders X cost of one order = 28000 X65

we assume basic EOS model of inventory control Inventory leuel 1 time Average number copies in storage = 2 = x cost of one copy storage cost to store -XX0.50 20.50 x.

Tatal cost = ordering cost + storage cost screze 5 0(X) = 2.8000x 65 +.52 x To minimise tatal cost, we will use concept q maxima and menina q caleciler Taking first derivanine с(x)

(w) * -in), elix)(20000457 016. My elim): 20000265 04) 4 ous en cl(a) = 2000x65 (toys (1) c'ex)= 20000X65X61X+0.5 ох үnaximum бх тиіпитит, c'cm) = 0 -> 20000 x65x61) 82 +0.5=0

o 5 = 28000x 6522 0.5=280000 x 65 x² = 28000X65X2 6.5 ? = 7280000 z = IN 7200000 5 X = 17200000 as a cannot 'be negative]

5 2= 2 698- 15 ccn) will be minimum at х= 269 8. / 5 ccx)20 с. - 2 698. 5

Taking second derivative g.com. ("(n)= a cik) c'(n) = (28000865367) from c'(x)2 –20000865 € (33)+ c!() = - 28000 x 65XC-2)X? to

clien) = 2800086582 x3 at x = 2698.15 d/12698.15) = 20000 8656421 (2698./537 So, don) to at 2=2698.15 So, X= 2698.15 au 2 698. are optimum numbers of copies per order

(A) 2428 (B) 26981 (C) 3257 (D) 3816 so , (B) option is correct.