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Chapter 4 100

# Chapter 4 100 - 352 CHAPTER 4 APPLICATIONS OF...

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Unformatted text preview: 352 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 2 2 2 4 \$0" -w)=7‘ —4r29c2+4924 => T2\$2—\$4=T4—4T2m2+4x4 => 5w4‘5r2\$2+r4=0. This is a quadratic equation in \$2. By the quadratic formula 3:2 = 5 3:0 5r2. but we reject the root with the + sign since it doesn‘t satisfy (*)1 [The right side is negative and the left side is positive] So ac : sﬁi 7” Since 5(0) 2 S(r )— a 0 the maximum surface area occurs at the critical number and \$2 =4 2 => y2 = r2 — 5 ’ 8/5? 2M7? => the surface area is W 1 >7" mm? ”122.5114: WIHC] 7er [Lt—5 ﬁ"—2‘2‘/5] :7r1"2 [i'E—f] :7r'r 2(1+\/_) 23. Perimeter230 :5 2y+w+7r<§> :30 :5 1 7m: a: 7m — 2 (30 It 2 ) — 15 — 2 I. The area is the area of the 2 rectangle plus the area of the semicircle. or my + \$743) . so A(m) — m(153E * E i 1:12) + in? 2 15\$ — @102 ~ £232. 1 . . A'(:1:) — 15 (1:1)3: — 0 w — 1 +:/4 — 46317. A”(:1:): i (1 + g) < 0.50th1sg1vesa . . , 60 30 157r 60 + 157r — 30 — 157r 30 max1mumi The d1mens10ns are a: : — — = —-——-—— 2 ft, 4+w 4+W 4+W 4+W so the height of the rectangle is half the base. 29. my : 384 :> y : 384/1 Total area is A(33) : (8 —I- \$)(12 + 384/510) 2 12(40 + :8 + 256/56), so A’(a:) : 12(1 7 256/1?) 2 0 => 3: : 16. There is an absolute minimum when a: : 16 since A’(ac) < 0 for 0 < a: < 16 and A’(a:) > 0 for a: > 16. When a: = 16. y : 384/16 : 24. so the dimensions are 24 cm and 36 cm. 30. my : 180. so y = 180/m. The printed area is (a: i 2)(y i 3) : (a: — 2)(180/:r ~ 3): 186 — 3:5 — 360/:E : A(w). A'(a:) : ,3 + 360/:n2 : 0 when \$2 : 120 => :c = 2%. This gives an absolute maximum since A'(:I:) > O for 0 < a: < 2 \/ 30 and A'(at) < 0 for cc > 2 m. When a: : 2 «30. y : 180/(2 1/30 ), so the dimensions are 2 \/ 30 in. and 90/ \/ 30 in. ...
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