AP2: Applied Optimization
Michael Nguyen – 03/01/2024
1. Problem introduction:
A water treatment plant is located on the north bank of a straight river 2 km wide. A pipeline will
be constructed from the plant to storage tanks on the south bank of the river, 6 km east of the
refinery. The cost of laying the pipe is $400,000/km over land to a point P on the north bank and
$800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should P be
located?
2. Problem illustration:
Figure 1 below illustrates the problem when sketching practically:
Figure 1. Demonstration of the pipeline implementation from the manufacturing plant to the storage tank
Note:
G_price represents the cost of laying the pipeline on the ground.
U_price illustrates the cost of constructing the pipeline underwater.
* = multiplication
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3. Solution:
Let x represent the distance (km) between the water treatment plant at the point P (AP), where pipe
enter the water, and the location from location P to the Storage Tank is called (PB), which is
illustrated in Figure 2 below:
Figure 2. Setting variables to solve the problem
According to the requirement, we have to find the optimized location of P to minimize the cost of
constructing the pipeline between the water treatment plant and the storage tank. The cost of
installing the pipeline can be calculated as follows:
T = Total cost = G_price * AP + U_price * PB ($) (Figure 2)
We have x represented the distance from the water treatment plant to location P, and the cost of
laying the pipeline on the ground is $400,000/km, which can be represented as:
G_price * AP = 400,000x ($)
The cost for laying the pipeline from location P to the storage bank (B) is complicated to calculate
without additional point. Therefore, I have marked point D that is the same distance from the
manufacture plant to location P, but it is located south of the river (Figure 2). When marking an
additional point into the problem, we can see that the distance from location P to the storage bank
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can be imagined as the hypotenuse of the 90° triangle, where PD and DB are the squared angles.
The distance DB demonstrates the remaining distance to the storage tank, which is 6 – x (km), and
PD is also the width of the river, which is 2 km. By applying the Pythagoras’s theorem into the
problem, we can calculate PB as follows:
PB2 = PD2 + DB2
= 22 + (6 – x)2
= 4 + 36 – 12x + x2
= x2 – 12x + 40
 PB = √𝒙𝟐 − 𝟏𝟐𝒙 + 𝟒𝟎
Therefore, the cost for laying PB can be seen as:
U_price x PB = 800,000 * √𝒙𝟐 − 𝟏𝟐𝒙 + 𝟒𝟎
Plug the PB and AP into the formula of total cost we have:
T = 400,000x + 800,000 * √𝒙𝟐 − 𝟏𝟐𝒙 + 𝟒𝟎
To find the minimum cost of the pipeline construction. First, we need to find the critical point the
derivative is equal to 0, which is also the point where it hits the maximum or minimum cost:
T’ = 400,000 + 400,000 *
-400,000 = 400,000*
-1 =
𝟐𝐱−𝟏𝟐
√𝒙𝟐 −𝟏𝟐𝒙+𝟒𝟎
=0
𝟐𝐱−𝟏𝟐
√𝒙𝟐 −𝟏𝟐𝒙+𝟒𝟎
𝟐𝐱−𝟏𝟐
√𝒙𝟐 −𝟏𝟐𝒙+𝟒𝟎
√𝒙𝟐 − 𝟏𝟐𝒙 + 𝟒𝟎 = 2x – 12
𝒙𝟐 − 𝟏𝟐𝒙 + 𝟒𝟎 = (2x – 12)2
x2 – 12x + 40 = 4x2 – 48x + 144
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 -3x2 + 36x – 104 = 0  x1 =
𝟏𝟖 − 𝟐√𝟑
𝟑
≈ 4.845, x2 =
𝟏𝟖 + 𝟐√𝟑
𝟑
≈ 7.1547
Although the function returns two critical numbers, only the result of 4.845 km satisfies the
requirement. Because x is the distance from the water treatment plant to location P, and location P
is located between the plant and the storage tank, the distance should be smaller between the plant
and the storage (6 km). The second result returned 7.1547 km, which is out of the range (0, 6) km.
However, at T’ = 0, there are two possible scenarios: the result can be maximum or minimum. For
that reason, it is necessary to conduct a first derivative test to validate the result:
x
T’ = 400,000 + 400,000 *
𝟐𝐱−𝟏𝟐
√𝒙𝟐 −𝟏𝟐𝒙+𝟒𝟎
4
𝟏𝟖 − 𝟐√𝟑
𝟑
5
-165685.4249
0
42229.1236
Table 1. First Derivative Test
The First Derivative Test shows that the T’ get negative when x = 4, and positive when x = 5. For
those reason, the total cost is decrease in the range (0,4.845) and hit the minimum value at x =
4.845 and then increase in the range (4.845, 6), which can be seen as the concave up type of chart
in general. Therefore, we can calculate the minimum total cost for implementing the pipeline from
water treatment plant to storage tank as below:
T = 400,000 * 4.845 + 800,000 * √(𝟒. 𝟖𝟒𝟓)𝟐 − 𝟏𝟐 ∗ 𝟒. 𝟖𝟒𝟓 + 𝟒𝟎
T = 3785640.646 ($)
The minimum cost for implement the pipeline is $3785640.646 when location P is at 4.845 km
away from the water treatment plant. For answer validation, Graphic illustrated by demos will be
include (link to the workspace: https://www.desmos.com/calculator/7sjejzmkpz):
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Figure 3. Answer validation with the Demos application.
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