Engineering Math ProjectBoat across the River

Supervised ByMr. yahya alskaji

2011 - 2012

Term 2

Lets start

Task 1 Students will search the web to find real life examples on working with vectors, e.g. adding forces, projectile, components, vectors in 3D, etc…

Vectors: Mathematics & Physics a quantity having direction as well as magnitude, esp. as determining the position of one point in space relative to another.

Task 2

If the boat is heading due north as it crosses a river with velocity 10 km/h of relative to water, the river has a uniform velocity of 5 km/h due to east. Graph the vectors in the chart below. Then determine the magnitude and direction of the boat’s velocity.

10km/h to northVelocity of the

boat 

5 km/h to eastVelocity of the river

 

The resultant velocity11.18 km/h and 63.4o above

the horizontal 

Task 2

Task 3 The boat is part of delivering company, it can hold 1200 kg and its capacity is 7.20 cubic meter. The service handles two types of boxes: small, which weight is up to 20 kg and no more than 1200 cm2, and large which 25 kg each and 1600 cm3 each.The delivery service charges 10Dhs for each small package, and 15 for each large package.Let : x be the number of small boxes

y be the number of big boxesa. Write an inequality to represent the weight of the packages in kg the boat can hold.

b. Write an inequality to represent the volume in cubic meter of packages the boat can hold.

Task 3

c. Graph the system of inequality

(0,48)

(0,0)(60,0)

d. Write a function that represents the amount of money the delivery service will make on each boat hold.

f(x,y) = 10 x + 15 y

Task 3

e. Find the number of each type of the boxes that should be placed on a boat to maximize revenue.

  To maximize the revenue they should deliver 0 small package and 48 large package

( 0 , 48 ) , ( 0 , 0 ) , ( 60 , 0 )

F (0,48) = 10(0) + 15(48) = 720$ F (0,0) = 10(0) + 15(0) = 0F (60,0) = 10(60) + 15(0) = 600$

f. What is the maximum revenue per boat?

F (0,48) = 10(0) + 15(48) = 720$ F (0,0) = 10(0) + 15(0) = 0F (60,0) = 10(60) + 15(0) = 600$

The maximum revenue is 720$

Task 4

Task 5 =121.24 N

W= 200 X 6 Sin 30= 600 J

Thank You For

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