A gardener with 800 m of available fencing wishes to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides as shown in the figure. Plot 1 Plot 2 What is the largest area that can be enclosed? (Use symbolic notation and fractions where needed.) Check the Student Hint for help starting this problem. A = m2A builder wishes to fence in 60000 m2 of land in a rectangular shape. For security reasons, the fence along the front part of the land will cost $40 per meter, while the fence for the other three sides will cost $20 per meter. How much of each type of fence should the builder buy to minimize the cost of the fence? 3) Determine the length of the fence along the front part of the land that will be cost $40 per meter. (Give your answer as a whole or exact number.) Check the Student Hint for help starting this problem. length: m b) Determine the length of the fence for the other three sides that will be cost $20 per meter. (Give your answer as a whole or exact number.) length: m 4\\ 'IT'I__'. 1. Alan _4_:_4:_4_d___- ___Ln c) What is the minimum cost? (Give your answer as a whole or exact number.) An closed box with a square base is to have a volume of 7400 cm'. What should the dimensions of the box be if the amount of material used is to be minimum? (Use decimal notation. Give your answers to three decimal places.) Recall: The surface area of a rectangular box is: SA = 2 . I . w + 2 . I . h + 2 . w . h, where / =length, w =width, and h = height. The volume of a box is: V = 1 . w . h. Hint: Check the Student Hint for help starting this problem. square base side length: cm height: cmA storage company designs a rectangular box with an open top that has a volume of 220 in'. Each box has a length that is three times its width. Calculate the minimum surface area of one of these boxes. Round your answer to three decimal places. Note: Draw a picture of the box and label the dimensions. Also, note the box here has no top. Volume: V = 1 . w . h Surface Area for a Box with no top: SA = 1 . w + 2 . w . h+ 2 . l . h Check the Student Hint for help starting this problem. minimum surface area: in 2A(X) = 35 555. 556 ma maximum is 35 555. 556 m