Question

QUESTIONS 13-18 PLEASE!

Maximum Storage Area DUE DATE: This project is worth 10% of your Unit 2 grade. Please review the Project FAQ handout for format and process. Problem Situation A construction company wishes to build a rectangular enclosure to store machinery and equipment. The site selected borders on a river that will be used as one of the sides of the rectangle. Fencing will be needed to form the other three sides. The company foot high chain-link fencing. The questions that follow should help you determine the dimensions of the rectangle that will produce the maximum area for the storage site. Assume that the company wishes to use all of the fencing, and that "width" refers to the measure of each of the two sides that are perpendicular to the river, and "length" refers to the measure of the side parallel to the river (see diagram. For the questions that follow, all measurement answers are to be given with the correct unit label, e.g. feet, square feet, etc has 682 feet of 10- width Storage Area width length PART 1 1. Copy the following table into your project and complete it, showing the dimensions (width and length) of some possible enclosures along with the resulting areas. (Recall that the company is using 682 feet of fencing total.) 225 Width (feet Length (feet Area (sq.ft) 25 632 50 582 75 100 200 250 275 15,800 29,100 2. Of all the dimensions (length and width) listed in the table, which choice gives the largest area for the enclosure? Express your answer with appropriate units 3. Let w represent the width, and / represent the length. Write an equation expressing / in terms of w 4. Write the general formula for the area A ofa rectangle in terms of I and w

Answer 1

Let the width of the storage area be w Let the length of the storage Area be l 2w + l = 682 Area (A) = l times w = w(682 - 2w) dA/dw = 682 - 4w = 0 rightdoublearrow w = 170.5 l = 341 d^2 A/dw^2 = -4 (Maxima) Thus Horizontal intercept will be 341 ft. & Vertical intercept will be 170.5 ft w times l = 50000 2w + l = 682 w(682 - 2w) = 50000 rightdoublearrow 2w^2 - 682w + 50000 = 0 w^2 - 341w + 25000 = 0 w = 341 plusminus squareroot (341)^2 - (4 times 25000)/2 = 341 plusminus 127.592/2 = 234.296 or 106.704 16. (i) w = 234.296: l = 213.408 (ii) w = 106.704: l = 468.592 Company can have following dimensions: (i) length of 213.408 feet & width of 234.296 feet (ii) length of 468.592 feet & width of 106.704 feet Solved above