Homework 8 [10/21/08]: Solutions
[2.6:4] (a) if x is the quantity per order and r is the number of orders placed during the year, then the
𝐼𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝐶𝑜𝑠𝑡 = 𝐼𝐶 = 𝑜𝑟𝑑𝑒𝑟𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 + 𝑐𝑎𝑟𝑟𝑦𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 = 160𝑟 + 16𝑥
𝑢𝑛𝑖𝑡𝑠
(b) The constraint function can be modeled by 𝑥 𝑜𝑟𝑑𝑒𝑟 ∙ 𝑟 𝑜𝑟𝑑𝑒𝑟𝑠 = 𝑥𝑟 𝑢𝑛𝑖𝑡𝑠 = 640 𝑢𝑛𝑖𝑡𝑠
(c) Minimize Inventory Cost :
𝐼𝐶 = 160𝑟 + 16𝑥 = 160𝑟 + 16
640
𝑟
= 106𝑟 + 16 640 𝑟 −1
𝐼𝐶 ′ 𝑟 = 160 − 16 640 𝑟 −2 = 160 1 − 64𝑟 −2
To minimize, set 𝐼𝐶 ′ 𝑟 = 0 and find that
64
1 = 𝑟 2 𝑜𝑟 𝑟 = ±8
We consider only the positive root
And proceed to find x:
𝑥𝑟 = 640; 8𝑥 = 640;
𝑥 = 80
So, ordering 80 units each of 8 times per year, inventory cost will be minimized
𝐼𝐶 = 160𝑟 + 16𝑥 = 160 8 + 16 80 = $2560
[2.6:6] (a) if x is the quantity per order and r is the number of production runs during the year, then the
5
𝐼𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 𝐶𝑜𝑠𝑡 = 𝑜𝑟𝑑𝑒𝑟𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 + 𝑐𝑎𝑟𝑟𝑦𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 = 15000𝑟 + 𝑥
2
We also know the constraint function can be modeled by 𝑥𝑟 = 600000
300000
So, 𝐼𝐶 𝑟 = 15000𝑟 + 5
and 𝐼𝐶 10 = 15000 10 + 5 30000 = 300000
𝑟
(b) Minimize Inventory Cost :
5
𝐼𝐶 = 15000𝑟 + 𝑥 = 15000𝑟 + 5 300000 𝑟 −1
2
𝐼𝐶 ′ 𝑟 = 15000 − 5 300000 𝑟 −2 = 15000 1 − 100𝑟 −2
To minimize, set 𝐼𝐶 ′ 𝑟 = 0 and find that
100
1 = 𝑟 2 𝑜𝑟 𝑟 = ±10
We consider only the positive root
And proceed to find x:
𝑥𝑟 = 600000; 10𝑥 = 600000;
𝑥 = 60000
So, ordering 60000 units each of 10 times per year [r = 10], inventory cost will be minimized
5
5
𝐼𝐶 = 15000𝑟 + 2 𝑥 = 15000 10 + 2 60000 = $300000
[2.6:8] We want to minimize 𝐼𝐶 = 40𝑟 + 2𝑥, where r denotes amount of times per year orders are
placed, and x denotes size of each order. We have the constraint 𝑥𝑟 = 8000. Note that this time, we
are dealing with ‘maximum’ inventory, not ‘average’ inventory, which leads to a slight amendment to
the IC.
𝐼𝐶(𝑟) = 40𝑟 + 2𝑥 = 40𝑟 + 2
8000
𝑟
= 40𝑟 + 16000𝑟 −1
𝐼𝐶 ′ 𝑟 = 40 − 16000𝑟 −2 = 40 1 − 400𝑟 −2
To minimize, 𝐼𝐶 ′ 𝑟 = 0 and find that
400
1 = 𝑟 2 𝑜𝑟 𝑟 = ±20
We consider only the positive root
And proceed to find x:
𝑥𝑟 = 8000;
20𝑥 = 8000;
𝑥 = 400
So, ordering 400 units each of 20 times per year [r = 20], inventory cost will be minimized
𝐼𝐶 = 40𝑟 + 2𝑥 = 40 20 + 2 400 = $1600
[2.6:18] The floor area of an x ft. by y ft. building must equal 12000 feet squared. Therefore, we have
our constraint 𝑥𝑦 = 12000. Further, we want to minimize our costs for putting up walls of such a
building, and cost can be modeled by 𝐶𝑜𝑠𝑡 = 70𝑦 + 2 50𝑥 + 50𝑦 where the side of the building with
glass wall has length y.
We are minimizing cost, 𝐶𝑜𝑠𝑡 = 120𝑦 + 100𝑥 subject to 𝑥𝑦 = 12000
𝐶𝑜𝑠𝑡 𝑦 = 120𝑦 + 100
12000
𝑦
= 120𝑦 + 100 12000 𝑦 −1
𝐶𝑜𝑠𝑡 ′ 𝑦 = 120 − 100 12000 𝑦 −2 = 120(1 − 10000𝑦 −2 )
To minimize, 𝐶𝑜𝑠𝑡 ′ 𝑦 = 0 and find that
10000
1 = 𝑦 2 𝑜𝑟 𝑦 = ±100
We consider only the positive root
And proceed to find x:
𝑥𝑦 = 12000;
100𝑥 = 12000;
𝑥 = 120
Therefore, a building with dimensions 120 ft. by 100 ft. [where the glass wall is 100ft long] will minimize
total cost
[2.6:20] Using the picture on page 197 of your book, the perimeter of the shape must be 440 yards.
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 2𝑕 + 𝜋𝑥 = 440. We want to use that as our constraint to maximize the area of the
rectangular portion of the field, or 𝑥𝑕.
We are maximizing area, 𝐴 = 𝑥𝑕 subject to 2𝑕 + 𝜋𝑥 = 440
440−𝜋𝑥
𝜋
𝐴 𝑥 =𝑥
= 220𝑥 − 2 𝑥 2
2
𝐴′ 𝑥 = 220 − 𝜋𝑥
To maximize, 𝐴′ 𝑥 = 0 and find that
𝟐𝟐𝟎
𝒙=
𝒚𝒂𝒓𝒅𝒔 and 𝑕 = 110 yards
𝝅
[2.6:22] Consider a rectangular box with base dimensions x and 2x, and height h.
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 = 27 = 2 𝑥𝑕 + 2 2𝑥 2 + 2 2𝑥𝑕 = 4𝑥 2 + 6𝑥𝑕
We want to maximize 𝑉 = 2𝑥 𝑥 𝑕 = 2𝑥 2 𝑕 = 2𝑥 2
𝑉 ′ 𝑥 = 9 − 4𝑥 2 , so setting this equal to zero,
9
4
27−4𝑥 2
6𝑥
2
=
𝑥
3
4
3
27 − 4𝑥 2 = 9𝑥 − 𝑥 3
3
= 𝑥 , 𝑜𝑟 𝑥 = ± 2. Let’s consider only the
positive root. The dimensions of the box, then, are
𝟑
, 𝟑, 𝟐
𝟐
𝒇𝒆𝒆𝒕.
[2.6:26] We want to find when the weekly sales amount is falling the fastest, or when the derivative at
its minimum.
1000
4000
𝑓 𝑡 = (𝑡+8) − 𝑡+8 2 = 1000 𝑡 + 8 −1 − 4000 𝑡 + 8 −2
𝑓 ′ 𝑡 = −1000 𝑡 + 8 −2 + 8000 𝑡 + 8 −3
𝑓 ′′ 𝑡 = 2000 𝑡 + 8 −3 − 24000 𝑡 + 8 −4 = 2000 𝑡 + 8 −3 1 − 12 𝑡 + 8 −1
Setting the second derivative equal to zero will give critical points of the graph of the derivative, which
will tell us about when the derivative is largest/smallest.
12
1 = 𝑡+8 𝒐𝒓 𝒕 = 𝟒 𝒘𝒆𝒆𝒌𝒔
That is the time that the rate is falling the fastest, or where the derivative f’(t) of the weekly sales f(t) is
minimized. Using test values on either side of 4, we see that f’’ is negative before t=4 and positive after
t=4. This tells us that f’ was decreasing then increasing, in other words reaches a minimum at 4.
Alternatively, we can find f’’’(t) to test concavity of f’(t) at t=4.
𝑓 ′′′ 𝑡 = −6000 𝑡 + 8 −4 + 4 24000 𝑡 + 8 −5
𝑓 ′′′ 4 = −6000 12 −4 + 4 24000 12 −5 > 0 so f’(t) is concave up at 4, indicating a
minimum at t=4
[2.7:2] 𝐶𝑜𝑠𝑡 = 𝐶 𝑥 = .0001𝑥 3 − .06𝑥 2 + 12𝑥 + 100
𝐶 ′ (𝑥) = .0003𝑥 2 − .12𝑥 + 12 = .0003 𝑥 2 − 400𝑥 + 40000 = .0003 𝑥 − 200 2
There is a critical point at 𝑥 = 200. However, the derivative is always positive, so the cost is
always increasing.
The graph of the marginal cost (C’(x)) is a parabola facing upward (“smiling”) with vertex at 200,
so before x=200, the graph is decreasing. By this logic, at x=100, the marginal cost is decreasing.
The minimum marginal cost, therefore, is at x=200. We could also use the second derivative
(C’’(x)) to find the critical points of the first derivative, resulting in the same conclusion.
[2.7:10] DRAW A PICTURE!!!  We want to minimize the area of a rectangle, 𝑎𝑏, subject to the
constraint 𝑏 = 30 − 𝑎. We get this constraint by realizing that the point (𝑎, 𝑏) must lie on the line
𝑦 = 30 − 𝑥.
𝐴𝑟𝑒𝑎 = 𝑎𝑏 = 𝑎 30 − 𝑎 = 30𝑎 − 𝑎2
𝐴𝑟𝑒𝑎′ = 30 − 2𝑎
𝐴𝑟𝑒𝑎′ = 0 = 30 − 2𝑎 ; 𝑎 = 15
Further, if 𝑎 = 15, 𝑏 = 30 − 15 = 15
The ordered pair (15,15) lies on the line 𝑦 = 30 − 𝑥, and when a rectangle is formed with (15,15)
according to the description in the book, the area of the rectangle formed is 225 𝑢𝑛𝑖𝑡𝑠 2
[2.7:12] When price is $50, 4000 tickets are sold. When price is $52, 3800 tickets are sold. Since we are
supposed to assume a linear demand curve, we can use the points (50,4000) and (52, 3800) to find that
𝑑𝑒𝑚𝑎𝑛𝑑 = 9000 − 100𝑥. We know, from the chapter, that 𝑅 𝑥 = 𝑥 ∙ 𝑑𝑒𝑚𝑎𝑛𝑑 = 9000𝑥 − 100𝑥 2
Minimizing 𝑅 𝑥 ,
𝑅 ′ 𝑥 = 9000 − 200𝑥 𝑖𝑠 0 𝑤𝑕𝑒𝑛 𝑥 = 45
So when ticket price is $45, revenue is maximized and is equal to $202,500
[2.7:14] Considering our problem context, we can use the line 𝑦 − 200 = −(𝑥 − 100) to represent the
amount brought into the club per person. We are looking for total revenue, so in a process similar to
the book’s treatment of demand and revenue, we will multiply the entire equation by x *memberships+.
This will also reconcile our units, since the current ordered pairs are (memberships, cost/membership).
This gives 𝐼𝑛𝑐𝑜𝑚𝑒 = −𝑥 2 + 300𝑥. Finding 𝐼𝑛𝑐𝑜𝑚𝑒 ′ 𝑥 = −2𝑥 + 300 and setting it equal to zero tells
us that when we sell 150 memberships, our income at the club will be maximized. This does not exceed
the total allowed memberships of 160.
[2.7:16] Considering the context of the problem, we can use the line 𝑦 − 36000 = −300(𝑥 − 100) to
represent the amount per car that contributes to the daily total toll collected [where x is in cents]. In
other words, when we increase the toll by 1 cent, we decrease the cars through the toll road by 300
[slope is -300]. Again, in an effort to reconcile our units, we multiply the entire equation by x cars,
finding that 𝐷𝑎𝑖𝑙𝑦 𝐼𝑛𝑐𝑜𝑚𝑒 = −300𝑥 2 + 66000𝑥. The derivative, or
𝐷𝑎𝑖𝑙𝑦 𝐼𝑛𝑐𝑜𝑚𝑒 ′ 𝑥 = −600 + 66000, set equal to zero tells us that when we charge 110 cents (or
$1.10), we will maximize the toll road’s revenue.
[2.7:18] 𝐷𝑒𝑚𝑎𝑛𝑑 = 200 − 3𝑥
𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑥 200 − 3𝑥 = 200𝑥 − 3𝑥 2
𝐶𝑜𝑠𝑡 = 75 + 80𝑥 − 𝑥 2
𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 − 𝐶𝑜𝑠𝑡
= 200𝑥 − 3𝑥 2 − (75 + 80𝑥 − 𝑥 2 )
= −2𝑥 2 + 120𝑥 − 75
′
𝑃𝑟𝑜𝑓𝑖𝑡 = −4𝑥 + 120
Profit has a critical point when 𝒙 = 𝟑𝟎, which when tested is shown to be a maximum.
[2.7:22] (a) 𝐶 60 = $1100
(b) 𝐶 ′ 40 = $12.50
(c) 𝐶 = $1200 𝑤𝑕𝑒𝑛 𝑥 = 100
(d) 𝐶 ′ = $22.50 𝑤𝑕𝑒𝑛 𝑥 = 20 𝑎𝑛𝑑 140
(e) 𝐶 ′ 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑖𝑧𝑒𝑑 𝑎𝑡 𝑥 = 80, 𝑤𝑕𝑒𝑟𝑒 𝐶 ′ = $5
1
[page 213:58] To put the problem in context, let’s consider the line 𝑦 − 25 = − 2 𝑥 − 40 which
models how the yield/tree changes as we change the amount of trees. We are looking for total yield, not
yield/tree, so in a process similar to the book’s, let’s multiply everything by x *trees+. This gives
𝑥2
𝑇𝑜𝑡𝑎𝑙 𝑌𝑖𝑒𝑙𝑑 = − + 45𝑥. We can maximize yield by taking its derivative, and 𝑌𝑖𝑒𝑙𝑑′ 𝑥 = −𝑥 + 45.
2
Setting this equal to zero, we find that our yield is maximized when we have 45 trees in our orchard.
Testing points near 45 will confirm that this is a max.
[page 213: 62] The total distance traveled can be modeled by 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 25 + 𝑥 2 + (15 − 𝑥)
Since we are to consider minimizing time, we need to think of our total time in the following way:
25+𝑥 2
15−𝑥
1
1
𝑇𝑖𝑚𝑒 = 8 + 17 = 8 25 + 𝑥 2 + 17 (15 − 𝑥). We don’t have issues with units, because x is in
miles, and our rates are mph.
𝑇𝑖𝑚𝑒 ′ 𝑥 =
0=
𝑥
8
25 + 𝑥 2
𝑥
25 + 𝑥 2
8
−1/2
−
8
= 𝑥 25 + 𝑥 2 −1/2
17
8 −2 25 + 𝑥 2
=
17
𝑥2
2
17
25
−1= 2
8
𝑥
25
2
𝑥 =
17 2
−1
8
𝑥=
25
17
8
2
=
−1
−1/2
−
1
.
17
To minimize time, we need to find when this derivative is zero.
1
17
25
=
289 64
−
64 64
25
=
225
64
64 8
=
9
3
[𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑜𝑛𝑙𝑦 𝑡𝑕𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑜𝑜𝑡]
𝟖
So, to minimize our time, we need to make 𝒙 = 𝟑 𝒎𝒊𝒍𝒆𝒔, which means that our total time will be
1
8
25 +
8
3
2
+
1
8
1 25 9
64 1 37
1 17 37 17 37
15 −
=
+
+
= ∙
+
=
+
𝑕𝑜𝑢𝑟𝑠
17
3
8
9
9 17 3
8 3 51 24 51
[which is a little less than 1.5 hours] to make our trip.