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Calculus: Related rates problem.

A multipurpose shoebox-shaped hall is planned. Its length is supposed to be
x = 50 meters, the width is supposed to be y = 30 meters, and the height is 10
meters. Furthermore, an expensive diagonal beam is planned, hosting a movable
camera car, starting in one (left, back, down) corner and going through the whole
hall until the (right, front, up) corner.

The cost of the base is $50 per square meter, the cost of the roof is $200 per
square meter, and the cost of the sides is $100 per square meter. Furthermore, one
meter length of the diagonal beam costs $ 3000.

a) Seeing the plan, the project leader wants the hall to be longer, maintaining the
height of 10 meters. Since the cost of the whole construction (including the diagonal
beam) is supposed to be constant, the width must be reduced. The question is: If
one increases the length x by 1 cm, how much must the width y be reduced to keep
the cost constant? More formally, the question is to find an equation between the rate
of change dx/dt of the length and the rate of change dy/dt of the width.

b) Can you find such an equation between dx/dt and dy/dt, not just for x = 50 and
y = 30, but for general x and y?

Tools required for this project:
Calculus: Related Rates
Algebra
Geometry: Theorem of Pythagoras

Solution Preview

Solution:
It will be helpful to first draw a picture of the hall and label the sides.

Determine the total cost of the original hall.
Total Cost = Cost of base + Cost of roof + Cost of 4 sides + Cost of beam
For example, Cost of base = area of base * price per square meter
= x*y*50
=50*30*5
=$75,000

Figure out the ...

Solution Summary

Steps to finding the adjusted dimensions of a shoebox-shaped hall and finding the related rates equation of the length and width of the hall. Includes a pdf attachment.

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