text: D(x) = 20x + 10 ≤ 130 - 20 = 110 (as there must be 20 ft above the present river surface\nSo 20x ≤ 100\nx ≤ 5\n\nW(x) = 10(x² - 8x + 22) ≤ 310\nie x² - 8x + 22 ≤ 31\nSo x² - 8x + 9 ≤ 0\nie (x - 9)(x + 1) ≤ 0\nSo -1 ≤ x ≤ 9\nAs x ≥ 0 (cannot have a negative depth or width)\n0 ≤ x ≤ 9\nSo to satisfy both conditions, 0 ≤ x ≤ 5\n\nWhat is maximum width in this range?\nWell minimum occurs when x = -b/(2a) = 8/2 = 4\nAt endpoints\nWhen x = 0 W = 220ft\nWhen x = 5 W = 70ft\nSo maximum width in this range is where the river is the shallowest.\nAlso minimum depth occurs at x = 0 when D = 10ft\nand maximum depth ocurs at x = 5 when D = 110ft\n\nSo now the choices need be made as the actual criteria being sought are not clearly spelt out.\n\nIn terms of straight economics it is probably wisest to have the almost least width and the maximum depth(at x = 5) but this is a guess.