Question

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C. 2. What rectangle has the smallest perimeter, for a given area? Suppose you want a rectangle with area 200. What choice of length L and width W will give the smallest perimeter? a. Sketch a rectangle and label it as having length L and width W. b. Write the area in terms of length and width: A = Write the perimeter in terms of length and width: P = d. If L = 200 and W = 1, calculate the area, and perimeter e. If L = 40 and W = 5, calculate the area, and perimeter f. If L = 20 and W = 10, calculate the area and perimeter 8. If L = 15, what does the width need to be to make the area 2007 W = This is a constraint. h. One cannot choose the length and width independently. To get an area of 200, write width W as a function of length L; W = This is how we solve the constraint. i. Write the perimeter in terms of just the length L: P(L) =. Now we have an objective function of one variable, and we can use calculus on it. 1. Use calculus to find the value of L that minimizes P(L). We can assume that .> 0 in this problem. 1. Calculate P'(L) =_ ii. Set P'(L) = 0 and solve for L to find any critical numbers. Write an exact number and a decimal approximation with 8 decimal places. iii. Find P"(L) to see if the critical number is a local minimum or maximum. iv. Examine intervals of increase and decrease to see if the critical number is an absolute minimum or just a local minimum V. Think about the graph of P(1) = 21+ *This function has a vertical asymptote at 2 - This function has a slant asymptote as L→ 00. Sketch the graph of P(L) as a function of L. All of this confirms that you have found an absolute minimum for the perimeter.

Answer 1

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