Solução_Calculo_Stewart_6e

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.