Engineering Math 1

Engineering Math 1
Engineering Math
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Engineering Math 2
Engineering Math
Question 1
Part a
Figure 1: Cable Laying
To connect the island to the nearest exchange, we need to calculate the length of the cable
required. Then, using the costs of the cable per unit length, we can get the total cost.
The distance AE=(42−x )km
Then, we can see that AI is the hypotenuse of triangle IPA. applying the Pythagoras Theorem,
AI2=252+ x2=625+ x2
AI is the distance under water hence, the length of the cable under water¿ AI= √625+ x2 km
Hence, the cost of the cable under water ¿ $ 3600 × √ 625+ x2=$ 3600 √ 625+ x2
AE is the distance on land. As a result, the length of the cable on land¿ ( 42−x ) km
Hence, the cost of the cable on land ¿ $ 2700 × ( 42−x ) =$ 2700 ( 42−x )

Engineering Math 3
Total cost of cable=Cost of cable under water + Cost of cable on land
Total Cable cost ,C=3600 √ 625+ x2 +2700 ( 42−x )
Part b
For the cost of laying the cable to be minimum, dC
dx =0
We know that, C=3600 √ 625+ x2+ 2700 ( 42−x )
dC
dx = d
dx ( 3600 √ 625+ x2 ) + d
dx ( 2700 ( 42−x ) )
¿ 3600 d
dx ( √ 625+ x2 ) +2700 d
dx ( 42−x )
Let 625+ x2=u so that d u
dx =2 x
√ 625+ x2= √ u=u
1
2
dC
du = 1
2 u
−1
2 = 1
2 √625+ x2
3600 dC
dx =3600 dC
du × du
dx =2 x 3600
2 √ 625+ x2 = 3600 x
√ 625+ x2
2700 d
dx ( 42−x ) =2700 ( −1 ) =−2700
dC
dx = 3600 x
√625+ x2 −2700=0
3600 x
√625+ x2 =2700