Math Task: Calculus Applications in Cost and Channel Optimization

This assignment explores the application of calculus in solving practical optimization problems. It consists of two main questions: The first question focuses on minimizing the cost of laying a cable both underwater and on land, using derivatives to find the optimal distance for the cable to transition from water to land. The solution involves setting up a cost function, differentiating it, and solving for the minimum cost. The second question deals with analyzing a channel profile, determining turning points, calculating the channel length using integration, and optimizing the channel's cross-sectional area to ensure safe flow capacity. The solution involves finding derivatives, evaluating integrals, and applying geometric principles to calculate areas and volumes. The assignment concludes by calculating the cost of the lining material required for the channel.

Math Task 1
Math Task
Student’s Name
Course
Professor’s Name
University
City (State)
Date

Math Task 2
Math Task
Question 1
Part a
To calculate the cost of laying the cable, we have to calculate the distances AE and AI.
AE=(42−x )
Then from Pythagoras Theorem, AI2=252+ x2=625+ x2
The length of the cable under water ¿ AI= √ 625+ x2 km
Hence, the cost of the cable under water ¿ $ 3600 √ 625+x2
The length of the cable on land¿ AE= ( 42−x ) km
Hence, the cost of the cable on land ¿ $ 2700 ( 42−x )
Total cost of cable=Cost of cable under water +Cost of cable on land
C (x)=$ ( 3600 √625+ x2 +2700 ( 42−x ) )
Part b
To get the minimum cost of laying the cable, dC ( x )
dx =0
dC ( x )
dx = d
dx $ ( 3600 √ 625+ x2 +2700 ( 42−x ) )
¿ 3600 d
dx ( √625+ x2 ) +2700 d
dx ( 42−x )
Let 625+ x2=u so that d u
dx =2 x

Math Task 3
√625+ x2= √u=u
1
2
dC ( x )
du =1
2 u
−1
2 = 1
2 √625+ x2
3600 dC ( x )
dx =3600 dC ( x )
du × du
dx =2 x 3600
2 √625+x2 = 3600 x
√ 625+ x2
2700 d
dx ( 42−x ) =2700 (−1 )=−2700
dC ( x )
dx = 3600 x
√ 625+ x2 −2700=0
3600 x
√625+ x2 =2700
(3600 x)2=27002(625+ x2)
625+ x2= ( 3600 ) 2
27002 x2= 16
9 x2
16
9 x2−x2=625
7
9 x2=625
x= √625 × 9
7 = √ 5625
7 = 75
√ 7 =28.3473 m
The cost of laying the cable is minimum when x=28.3473 m
Minimum cost¿ C (28.3473)=$ ( 3600 √625+28.34732+ 2700 ( 42−28.3473 ) )
Minimum cable cost=$ 172,929.4045

Math Task 4
Question 2
Part a
Evidently, there are 8 turning points in the proposed channel profile as shown in figure 1 in the
question paper. Four are maximum turning points while the other four are minimum turning
points. Given that, y=−2.014 sinx +1.922cosx−0.452 lnx−0.22 x+ 12.46, the turning are shown
in figure 1.
Figure 1: Turning points
Part b
y=−2.014 sin(x )+1.922 cos (x)−0.452 ln( x)−0.22 x +12.46
Length of the channel ¿ ∫
x=0.003
x=25
√ 1+( y ' (x))2 dx

Math Task 5
y' ( x ) = dy
dx = d
dx ( −2.014 sin ( x ) +1.922 cos ( x ) −0.452 ln ( x ) −0.22 x+12.46 )
¿−2.014 cos ( x )−1.922 sin(x )− 0.452
x −0.22
¿ y' ( x ) ∨¿2.014 cos ( x ) + 1.922sin ( x ) + 0.452
x +0.22
Channel length= ∫
x=0.003
x=25
√1+(2.014 cos ( x ) +1.922 sin ( x ) + 0.452
x +0.22)
2
dx
Solving the equation above we obtain, Channel length=55.41631km
Part c
Maximum channel width ¿ 6 m
Minimum cross section area ¿ 8 m2
Depth of channel=h
The x-intercepts are x=−3∧, x=3
Which implies that x +3=0∧, x−3=0
Then, ( x +3 ) ( x−3 )=0= y
Expanding the equation, we get,
y=x2−9
The equation of the shaded region is −h− ( x2−9 ) =9−h−x2
The curve y=x2−9 and the line y=−h meet at:

Math Task 6
y=x2−9=−h
x2−9=−h
x2=9−h
x=± √ 9−h
Area of shaded region ¿ 2 ∫
0
√ 9−h
(9−h−x2)dx ≥ 8 m2
2 [9 x−hx− x3
3 ]0
√9−h
=8
9 √ 9−h−h √ 9−h− ( √ 9−h)3
3 = 8
2 =4
(9−h) √ 9−h−(9−h) √ 9−h
3 =4

Math Task 7
(9−h) √9−h−(9−h) √9−h
3 =4
2
3 ( 9−h ) √ 9−h=4
( 9−h ) √9−h=4 × 3
2 =6
√ ( 9−h)((9−h)2 )=6
√( 9−h)3=(9−h)
3
2 =6
9−h=(6)
2
3
h=9− ( 6 )
2
3 =5.698 m
The depth of the channel below the ground level is ( 9−5.698 ) m=3.302 m
y=x2 +h−9=x2 +5.698−9=x2−3.302
The channel profile equation is y=x2−3.302
Part d
Channel height¿ 9 m
Safe flow capacity¿ 9 m× 0.75=6.75 m

Math Task 8
Safelevel , y = ( 6.75−3.302 ) m=3.448 m
But we know that the equation of the channel profile equation is y=x2−3.302
So, y=3.448=x2−3.302
x2=3.448+3.302=6.75
x= √6.75=±2.598
Maximum safe area ¿ 2 ∫
0
2.598
¿ ¿
¿ 2 ∫
0
2.598
(6.75−x2 ) dx
¿ 2 [ 6.75 x− x3
3 ]0
2.598
=2 ( 6.75 × 2.598− 2.5983
3 )=23.3827 m2
Max safe area=23.3827 m2
Part e

Math Task 9
Total area where the lining material will be used is the area of the channel up to ground level.
That is, ( 23.3827−8 ) m2=15.3827 m2
Cost of lining material Total area×Cost per unit area
Cost =15.3827 m2 × $ 30/m2
Cost =$ 461.481