Solução_Calculo_Stewart_6e

F.
200 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
51. Every line segment in the first quadrant passing through (a, b) with endpoints on the x-
and y-axes satisfies an equation of the form y − b = m(x − a),wherem 0 and m− 3 b
.Thus,S has its absolute minimum value when m = − 3 b
a
That value is
S
− 3
b
a
2 2
=
a + b 3 a
+
−a 3 b
− b =
a + 3√ 2
3√ 2
ab
b a 2 + a2 b + b
= a 2 +2a 4/3 b 2/3 + a 2/3 b 4/3 + a 4/3 b 2/3 +2a 2/3 b 4/3 + b 2 = a 2 +3a 4/3 b 2/3 +3a 2/3 b 4/3 + b 2
The last expression is of the form x 3 +3x 2 y +3xy 2 + y 3 [=(x + y) 3 ] withx = a 2/3 and y = b 2/3 ,
so we can write it as (a 2/3 + b 2/3 ) 3 and the shortest such line segment has length √ S =(a 2/3 + b 2/3 ) 3/2 .
53. (a) If c(x) = C(x)
x
, then, by Quotient Rule, we have c0 (x) = xC0 (x) − C(x)
.Nowc 0 (x) =0when xC 0 (x) − C(x) =0
x 2
and this gives C 0 (x) = C(x)
x
= c(x). Therefore, the marginal cost equals the average cost.
(b) (i) C(x) =16,000 + 200x +4x 3/2 , C(1000) = 16,000 + 200,000 + 40,000 √ 10 ≈ 216,000 + 126,491, so
C(1000) ≈ $342,491. c(x) =C(x)/x = 16,000
x
C 0 (1000) = 200 + 60 √ 10 ≈ $389.74/unit.
(ii) We must have C 0 (x) =c(x) ⇔ 200 + 6x 1/2 = 16,000
x
x =(8,000) 2/3 = 400 units. To check that this is a minimum, we calculate
+200+4x 1/2 , c(1000) ≈ $342.49/unit. C 0 (x) = 200 + 6x 1/2 ,
+ 200 + 4x 1/2 ⇔ 2x 3/2 =16,000 ⇔
c 0 (x) = −16,000
x 2 + 2 √
x
= 2 x 2 (x3/2 − 8000).Thisisnegativeforx400,soc is decreasing on (0, 400) andincreasingon(400, ∞). Thus,c has an absolute minimum
at x = 400. [Note: c 00 (x) is not positive for all x>0.]
(iii) The minimum average cost is c(400) = 40 + 200 + 80 = $320/unit.
a .