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Chapter 4 106

# Chapter 4 106 - 368 CHAPTER 4 APPLICATIONS OF...

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Unformatted text preview: 368 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 49. Note that IADI = IAP| + [PD] => 5 = in + [PDI => |PD| = 5 — :17. Using the Pythagorean Theorem for APDB and APDC gives us L(m):|AP|+ IBP|+ICP| :ac+ i/(5—m)2+22+ (5—33)2+32 =m+\/:c2—10:c+29+ \$2—10\$+34 m—5 + (8—5 s/mQ ~ 10x+29 «9:2 ~ 10w+34' 12 0.3 - 3 l‘ 4 0 5 9 —0.3 From the graphs of L and L’, it seems that the minimum value of L is about L(3.59) : 9.35 m. :> L’(a:) 2 1+ 50. We note that since 0 is the consumption in gallons per hour. and v is the velocity in miles per hour, then c _ gallons/hour # gallons 2} miles/hour mile gives us the consumption in gallons per mile. that is. the quantity G. ‘To ﬁnd the dc d1} dc .. dG d ”——C— 0—” minimum. we calculate E : %(g) : —d—v—;§—dl * —d:—2 This is 0 when 1} E — c = 0 4:) E : E. This implies that the d1} d1) 71 tangent line of C(v) passes through the origin. and this occurs when 1) z 53 mi/h. Note that the slope of the secant line through the origin and a point (2). C(11)) on the graph is equal to 0(1)). and it is intuitively clear that G is minimized in the case where the secant is in fact a tangent. 51. A The total time is a T(z): (time fromAto C) + (time fromCto B) C d_x / 2 2 /2 _ 2 x :_a—H_+_b_j—ML~ 0<\$<d U1 U2 [7 , m d—m sin61 Sin62 B Tva‘m‘T—Tz I«——d———-| 1 2 . . Sin 01 sin 62 The minimum occurs when T/(ZL’) = 0 :> = . 01 112 [Notes T"(x) > 0] ...
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