## Answers

perimeten 2 Square in à type of rectangle which has the smallest perimeter for a given area т. the length L and width. w has to have smallest perimeter ... Length (2) width(t) 5200 i 200 area to be salne A = Length x Width = L XA area = 2000 1 = 200 2 c) Perimeter 2 (L+A) d) Ib L = 200 and we and perimeter (200 +1) 402 e) Ib L=40 and w=5, area = 40 x 5 22.00 and 2(40+50) - 180 If L = 20 and wr 10 then area= 20x10 = 200 and perimeten 2 (10+20) = 60 then. W needs to be 40 B ) and A 200 If = 15 and 200 15 3

h ) A 200 We com know that L.

WFA - > LXW 200 200 Now perimeten in teams of length will be P(L) = 2 (WHL) 2 (20+) 400 + 26² L j i> p (L) 400+22² 7 d. dL 04224) - 4.42 - (48012). L 21-400 2 i) pikk) 50 21- 400 => LP 244 = 400 7) 2 = 200 L = 5200 [as Lyo we won't take La-20) 14. 14213562

iii) p"(L) 22- 400 d de L² [21 de (2- 400 d FO-480. (2).

.
? 800 <3 Now p" (1200) 800 (1200) So, the critical number 1200 h a local mininum We have got Io $200 14.14215562 Lett take avalue less than 5200 2*10% +400 P (10) = leb tane 10 then 10 and take let tane 15 a number greater that $200 2x152 +400 then p(15) 56.667 15 Now, P (520) P 5.6. 588 L>0 P(10) > PCF260) < P(15) and it na global minima. For vertical asympto je the de nominator P(L)=0 9 L20

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