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Please answer all C. 2. What rectangle has the smallest perimeter, for a given area? Suppose...

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please answer all
C. 2. What rectangle has the smallest perimeter, for a given area? Suppose you want a rectangle with area 200. What choice of
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.

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 hash ) 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 Liii) p(L) 22- 400 d de L² [21 de (2- 400 d FO-480. (2).</p><p>? 800 <3 Now p (1200) 800 (1200) So, the critical number 1200 h a l

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