Thomas Calculus 13th [Solutions]

Section 4.6 Applied Optimization 301
53. (a)
(b)
A( q) 1 h
2
2
kmq cm q , where 0 ( )
h hq km
3
q A q kmq and A ( q) 2 kmq . The critical
2
2 2
2q
points are 2 km , 0,
h
km
h
, but only 2km is in the domain. Then A 2km
0 at q 2km there
h
h
h
is a minimum average weekly cost.
A( q) ( k bq) m 1
cm h q kmq bm cm h q , where q 0 A ( q ) 0 at q 2km as in (a). Also
q
2 2
h
A ( q) 3
2kmq 0 so the most economical quantity to order is still q 2km which minimizes the
h
average weekly cost.
2
54. We start with c( x ) the cost of producing x items, x 0, and c( x)
the average cost of producing x items,
x
assumed to be differentiable. If the average cost can be minimized, it will be at a production level at which
d c( x)
xc ( x) c( x)
0
0 (by the quotient rule) xc ( x) c( x ) 0 (multiply both sides by
dx x 2
x
2 )
x
c( x)
c ( x ) where c ( x ) is the marginal cost. This concludes the proof. (Note: The theorem does not assure a
x
production level that will give a minimum cost, but rather, it indicates where to look to see if there is one. Find
the production levels where the average cost equals the marginal cost, then check to see if any of them give a
minimum.)
55. The profit p( x) r( x) c( x) 6 x
3
( x
2
6x 15 x) 3
x
2
6x 9 x , where x 0. Then p ( x) 2
3x 12x
9
3( x 3)( x 1) and p ( x) 6x 12. The critical points are 1 and 3. Thus p (1) 6 0 at x 1 there is a
local minimum, and p (3) 6 0 at x 3 there is a local maximum. But p (3) 0 the best you can do is
break even.
c( x) 2
56. The average cost of producing x items is c ( x) x 20x 20,000 c ( x) 2x 20 0 x 10, the
x
only critical value. The average cost is c (10) $19,900 per item is a minimum cost because c (10) 2 0.
57. Let x the length of a side of the square base of the box and h the height of the box. V x 2 h 48 h 48 .
2
x
3
The total cost is given by C 2 2 2
6 x 4(4 xh ) 6 x 16 x 48 6 x 768 , 0 12 768 12x
768
2 x
x C x 2 2
x x x
3
0 12x
768
3
C
0 12x 768 0 x 4; C 12 1536 C (4) 12 1536 0 local minimum.
2
2
2
x
x
4
48
2
x 4 h 3 and C (4) 6(4) 768 288 the box is 4 ft 4 ft 3 ft, with a minimum cost of $288.
2
4
4
58. Let x the number of $10 increases in the charge per room, then price per room 50 10 x , and the number of
rooms filled each night 800 40x the total revenue is R( x) (50 10 x)(800 40 x)
2
400x 6000x 40000, 0 x 20 R ( x) 800x 6000; R ( x) 0 800x
6000 0
x 15
2 ; R ( x) 800 R 15 800 0 local maximum. The price per room is 50 10 15 $125.
2
2
2
59. We have dR
dM CM M . Solving 2
d R
dM
maximum.
3
C 2M 0 M C . Also. d 2 0
R3
dM
2 2
at M C there is a
2
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