MATH104/184
Business Optimization Problems
MATH 104/184
November 12, 2015
1. A fuel tank of volume 144π cubic meters consists of a cylindrical part
capped by hemispheres at each end. The construction cost is $5 per
square meter for the cylindrical wall, but $10 per square meter for the
hemisphere ends. Find the minimum total cost for constructing such a
tank.
SOLUTION
SA of Cylinder = 2πrh
SA of Sphere = 4πr2
3
V of Sphere = 4πr
3
V of Cylinder = πr2 h
Cost of constructing the tank will be
C(r, h) = 5(2πrh) + 10(4πr2 ).
(1)
Since the cost function is a function of r and h, we want to make it
into only a function of one variable. We know that the volume of the
whole tank is 144π so we say:
4πr3
= 144π.
3
Isolating for h is easier so we will do that.
πr2 h +
4r3
= 144
3
(3)
144 4r
−
r2
3
(4)
r2 h +
h=
(2)
1
MATH104/184
Substituting (4) into (1), we will get:
144 4r
+ 10(4πr2 ).
C(r) = 5 2πr
−
r2
3
(5)
After simplification, we will have
C(r) =
1440π 40πr2
−
+ 40πr2 .
r
3
(6)
Take the derivative and setting = 0, we get
−1440π 80πr
−
+ 80πr = 0.
r2
3
(7)
Now we solve for r. Multiply both sides by r2 to make the equation
easier to solve.
80πr3
+ 80πr3 = 0
(8)
−1440π −
3
160πr3
= 1440
3
r3 = 27
(10)
r=3
(11)
(9)
Plugging r = 3 into (6) we will get:
C(3) =
1440π 40π32
−
+ 40π32
3
3
(12)
C(3) = $2261.95
(13)
2. Fion is an 80 year old widow and is diagnosed with lung cancer. Her
doctors say she will probably die one year from today. She currently
owns an apartment that is worth $50 million and it is expected that the
building will worth $45 million one year from today. When she dies,
she will donate all her assets (cash and apartment) to the BC Cancer
Society. Fion is not sure whether she should sell the building now or
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MATH104/184
donate the building property in one year. Her annual profit from collecting rent is given by the function
P (x) = −96x2 + 38, 400x − 960, 000.
(14)
Where x is the number of rooms rented out.
There are only 250 rooms in the apartment.
(a) How many rooms should Fion rent out to maximize profit?
(b) What is the maximum profit?
(c) Should Fion sell her property today, or wait until one year from
now to donate the $45 million property plus the rent profit from
the past year to the BC Cancer Society?
(Assume rents are collected at the end of each year)
SOLUTION
(a)
P 0 (x) = −192x + 38, 400 = 0
(15)
x = 200
(16)
(b) Need to test out end points as well as answer from a.
We know that 0 ≤ x ≤ 250.
P (0) = −960, 000
P (200) = 2, 880, 000
P (250) = 2, 640, 000
Therefore, the maximum profit is $2,880,000.
(c) If Fion decides not to sell the apartment today, she will be donating
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MATH104/184
$45 million +$2,880,000 = $47,880,000.
Therefore, it would be wise to sell the apartment today for $50
million and accumulate interest for the last year of her life before
donating it to charity.
3. A gas pipeline is to be constructed from a storage tank, which is right
on a road, to a house which is 600 feet down the road and 300 feet back
from the road. Pipe laid along the road costs $8.00 per foot, while pipe
laid off the road costs $10.00 per foot. What is the minimum cost for
which this pipleline can be built?
(Assume the pipeline path is piecewise linear, with at most two pieces.)
SOLUTION
Figure 1: This is the diagram.
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MATH104/184
√
C(x) = 8(600 − x) + 10 3002 + x2
√
C(x) = 4800 − 8x + 10 3002 + x2
C 0 (x) = −8 + √
5 · 2x
=0
3002 + x2
(17)
(18)
(19)
√
8 3002 + x2 = 10x
(20)
64(3002 + x2 ) = 100x2
(21)
x = 400
(22)
Now we need to find C(400) which is the substituition of 400 into x
into equation (17)
C(400) = $6600.
(23)
4. Kevin and his friends recently started a company selling calculators.
The start-up cost is $50,000 and the cost of each calculator is $5.50.
A recent market study shows if the price, p, is $5.50, the quantities
sold will be 150,000 units, but for every increase of $5 of the price,
the quantities sold will be halved. What should price p be in order to
maximize profit.
SOLUTION
We all know that
P =R−C
and
R = p · q.
We need to find the function of profit as a function of price p. To find
quantity q as a function of p, we need some knowledge of half-life.
p−5.5
5
1
q = 150, 000
2
(24)
If p is $5.50, then q would be 150,000 and for every $5 increase in price,
the quantities sold will be halved. (e.g. if p is $10.50 then 150,000 will
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be halved once.) The revenue function will be:
p−5.5
5
1
.
R(p) = 150, 000 · p ·
2
(25)
To find the cost function, separate it into variable costs and fixed costs.
Therefore,
z
}|
{
p−5.5
5
z }| {
1
.
C(p) = 50, 000 + 5.50 · 150, 000
2
(26)
Profit will be:
p−5.5
p−5.5
5
5
1
1
P (p) = 150, 000 · p ·
− 50, 000 − 5.50 · 150, 000
. (27)
2
2
p−5.5
5
1
(p − 5.50) − 50, 000
P (p) = 150, 000
2
(28)
" p−5.5
p−5.5
#
5
5
1
1
1
∂P
1
= 150, 000
+ (p − 5.50)
· ln
·
(29)
∂p
2
2
2
5
∂P
= 150, 000
∂p
150, 000( 12 )
p−5.5
5
p−5.5
5
1
p − 5.50
1
· 1+
· ln
=0
2
5
2
(30)
will always be positive so that part can be ignored.
p − 5.50
1
1+
· ln
=0
(31)
5
2
p − 5.50 =
p = 5.50 +
6
5
ln 2
−5
ln 12
≈ $12.71
(32)
(33)