chapter 1 8 - 8 CHAPTER 1 FUNCTIONS AND MODELS 49 Let the...

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Unformatted text preview: 8 CHAPTER 1 FUNCTIONS AND MODELS 49. Let the length of a side of the equilateral triangle be m. Then by the Pythagorean Theorem, the height 3; of the . . 2 trlangle satisfies y2 + (£95) 2 x2, so that y2 = $2 — im2 = 3.12 and y 2 gm Using the formula for the area A of a triangle. A 2 %(base)(height), we obtain A(a:) 2 fix) (gm) = $152, with domain a: > 0. 50. Let the volume of the cube be V and the length of an edge be L. Then V = L3 so L = W, and the surface area is 2 S(V) = 6 (W) = 6V2/3, with domain V > 0. 51. Let each side of the base of the box have length m, and let the height of the box be h. Since the volume is 2, we know that 2 = hm2, so that h = 2/ar2, and the surface area is S : m2 + 4mh. Thus, 3(a)) : m2 + 431(2/x2) = 3:2 + (8/m), with domain I > 0. 2 52. The area of the window is A : ash + -;-7r(%:v)2 = xh + %, where h is the height of the rectangular portion of the window. The perimeter is P : 2h + :1: + énat : 30 <=> 2h : 30 — $ — %7m (1) h = %(60 — 2x — my). Thus, 60 — 2m — 7m: 7T$2 A3: =a:————+——:15w—1302—£$2+1$2=15m-3x2—£a:2:15m—a:2 5L4 4 8 2 4 8 8 s 3 Since the lengths x and h must be positive quantities, we have x > 0 and h > 0. For h > 0, we have 2h > 0 c) 60 . Hence, the domain of A is 0 < m < 60 2 + 7r 2 + 71" 53. The height of the box is a: and the length and width are L = 20 — 2:0, W = 12 — 21:. Then V = LWac and so V(:c) = (20 — 2m)(12 — 2x) (:12) 2 4(10 4 ac)(6 — m)($) : 4:3(60 — 162: + m2) = 4:33 — 64:02 + 2409:. The sides L, W, and 20 must be positive. Thus, L > 0 4:) 20 — 2:5 > 0 <=> cc < 10; W > 0 4:) 30—m—é7ra3>0 <=> 60>2w+7rm 4:) ac< 12 — 2:1: > 0 4:) a: < 6; and w > 0. Combining these restrictions gives us the domain 0 < cc < 6. 54' $2.00 if 0.0 < a: g 1.0 2.20 if 1.0 < :1: g 1.1 2.40 if 1.1 < x S 1.2 2.60 if 1.2 < :1: g 1.3 2.80 if 1.3 < m S 1.4 3.00 if 1.4 < :1: S 1.5 3.20 if 1.5 < m g 1.6 3.40 if 1.6 < a: S 1.7 3.60 if 1.7 < a: g 1.8 3.80 if 1.8 < a: S 1.9 4.00 if 1.9 < x < 2.0 II C (cc) (b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400. On $26,000, tax is assessed on $16,000, and 10%($10,000) + 15%($6000) = $1000 + $900 = $1900. 55. (a) R070 15 10 20,000 I (in dollars) ...
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