Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 4.7 OPTIMIZATION PROBLEMS ¤ 201 55. (a) We are given that the demand function p is linear and p(27,000) = 10, p(33,000) = 8, so the slope is 10 − 8 = − 1 and an equation of the line is y − 10 = − 1 27,000 − 33,000 3000 3000 (x − 27,000) ⇒ y = p(x) =− 1 3000 x +19=19− (x/3000). (b) The revenue is R(x) =xp(x) =19x − (x 2 /3000) ⇒ R 0 (x) =19− (x/1500) = 0 when x =28,500. Since R 00 (x) =−1/1500 < 0, the maximum revenue occurs when x =28,500 ⇒ the price is p(28,500) = $9.50. 57. (a) As in Example 6, we see that the demand function p is linear. We are given that p(1000) = 450 and deduce that p(1100) = 440,sincea$10 reduction in price increases sales by 100 per week. The slope for p is so an equation is p − 450 = − 1 1 10 (x − 1000) or p(x) =− x +550. 10 (b) R(x) =xp(x) =− 1 10 x2 +550x. R 0 (x) =− 1 x + 550 = 0 when x = 5(550) = 2750. 5 p(2750) = 275, so the rebate should be 450 − 275 = $175. 440 − 450 1100 − 1000 = − 1 10 , (c) C(x) =68,000 + 150x ⇒ P (x) =R(x) − C(x) =− 1 10 x2 + 550x − 68,000 − 150x = − 1 10 x2 +400x − 68,000, P 0 (x) =− 1 x +400=0 when x =2000. p(2000) = 350. Therefore, the rebate to maximize profits should be 5 450 − 350 = $100. 59. Here s 2 = h 2 + b 2 /4,soh 2 = s 2 − b 2 /4. TheareaisA = 1 2 b s 2 − b 2 /4. Let the perimeter be p,so2s + b = p or s =(p − b)/2 ⇒ A(b) = 1 b (p − b) 2 2 /4 − b 2 /4=b p 2 − 2pb/4. Now A 0 p2 − 2pb bp/4 (b) = − 4 p2 − 2pb = −3pb + p2 4 p 2 − 2pb . Therefore, A 0 (b) =0 ⇒ −3pb + p 2 =0 ⇒ b = p/3. SinceA 0 (b) > 0 for bp/3,there is an absolute maximum when b = p/3.Butthen2s + p/3 =p,sos = p/3 ⇒ s = b ⇒ the triangle is equilateral. 61. Note that |AD| = |AP | + |PD| ⇒ 5=x + |PD| ⇒ |PD| =5− x. Using the Pythagorean Theorem for ∆PDB and ∆PDC gives us L(x)=|AP | + |BP| + |CP| = x + (5 − x) 2 +2 2 + (5 − x) 2 +3 2 = x + √ x 2 − 10x +29+ √ x 2 − 10x +34 ⇒ L 0 (x) =1+ x − 5 √ x2 − 10x +29 + x − 5 √ . From the graphs of L x2 − 10x +34 and L 0 , it seems that the minimum value of L is about L(3.59) = 9.35 m.
F. TX.10 202 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION √ 3=tan ψ + π ⇒ ψ = π . 6 6 63. The total time is T (x)=(time from A to C)+(time from C to B) √ a2 + x = b2 +(d − x) + , 0 0 for 0 ≤ t< √ 1 3 and f 0 (t) < 0 for t> √ 1 3 ,sof has an absolute maximum when t = √ 1 3 and tan θ = 2 1/ √ 3 1+3 1/ √ 3 2 = √ 1 3 ⇒ θ = π 6 . Substituting for t and θ in 3t =tan(ψ + θ) gives us
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