Optimization Module Introduction

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Optimization ModuleIntroduction
Optimization problems can sometimes take a toll
on Calculus students. This exercise can get them
out and about to see their knowledge put to use.
The problem entails them to seek out the longest
possible rod to carry around a corner in a
building. It is more fruitful to seek out a
corner in the Educational institution that you
are at. Try getting them involved
taking measurements, etc. If time allows, you can
cut rods of their proposed lengths and they can
see their own outcomes come to life! If not, you
can still use this presentation as a nice
walk-through.
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The Problem
Suppose the college is trying to order a
pipe that needs to be installed in a particular
room that is around a narrow corridor that meets
at a corner. Ordering two smaller pipes is out of
the question, as one long rod can be
ordered much cheaper. So, the college has
employed you some of their own prized Calculus
students to recommend to them the length of
the longest possible rod for the room.
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Getting Started
Solve a simpler problem first. Even though the
rod can be lifted off the ground to be the
longest, first find the longest rod that can be
slid around the corner on the floor. Let us
assume that the rod has no width.
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The Measurements
Suppose youve measured
W1 129.5 cm W2 106.7 cm H 335.3 cm
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Creating the Equation
Since we are trying to solve the simpler
problem first, we can think of the two
dimensional model
The length of the rod can be broken down to the
sum of the two lengths, L1 and L2
? Max L L1 L2
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Finding Substitutions
We try to look to find a common variable that we
can substitute for L1 and L2. Try ?.
Thus,
sin ? W2 / L2 cos ? W1 / L1
Or,
L2 W2 / sin ? L1 W1 / cos ?
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Finding the maximum
? Max L W1 sec ? W2 csc ?
L W1 sec ? tan ? W2 cot ? csc ?
Setting L 0 using either symbolic
calculator or a CAS like Mathematica, we see
? 0.75 radians
Thus, the longest rod will be, L(0.75)
333.5cm.
Q How do you know this is a max?
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Finding the width
Suppose then, you call the place where the
college will order the rod, and find that
the rod will have diameter measuring 6cm.
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Accounting for the width
If we still think in the two dimensions, we
can calculate the length of rod with width 6cm.
From this, we can see how to cut the pipe. Let us
blow up the area of concern
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Cutting the excess length
From trigonometry, we see the length ltop we wish
to remove is found by
tan ? 6/ltop
ltop 6.44 cm
tan(?/2-?) lbottom/6
lbottom 6.44 cm
Click for more help on how to calculate these
values.
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Length of Rod on the Floor
Therefore, we need to cut
ltop lbottom 6.44 6.44 12.88 cm.

• The length of the longest rod to slide
• along the floor is

L 333.5cm 12.9 cm
320.6cm
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Lifting the rod
If we lift the rod, we can fit a longer
rod through the corner. To find this, we
simply need to apply the Pythagorean Theorem.
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Pythagorean Theorem
The sum of the squares of the legs is equal
to the square of the hypotenuse.
L2 H2 C2
320.62 335.32 C2
463.9 C
Do we need to take the width of the rod into
consideration for the top and bottom?
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The Angle to the Floor
In much of the same way we needed to trim a
little of the rod due to the width, we may also
need to do a little trimming again
Using the length of the pipe just calculated, and
the measurement of the height, we see
tan ? H/L 335.3 / 320.6 ? .81 radians
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Making the Final Cut
Since we now have the angle where the pipe meets
the floor, we can use the similar ideas of
cutting this extra length as we did when we
found the length sliding across the floor.
tan ? 6/ltop
ltop 5.71 cm
and,
tan 90-? lbottom/6
lbottom 5.71 cm
? The longest rod will be 463.9 11.4 452.5cm
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Appendix More on cutting
If you are having difficulty trying to
determine the excess length we need to cut off,
this appendix is just what you seek!
On the floor
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Appendix (cont.)
Step 1. Draw straight edges where the pipe
will be cut
Step 2. Mark angle ? which you have previously
found.
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Appendix (cont.)
Step 3. If we assume walls of the hall
are parallel, from geometry, we know that if a
straight line cuts two parallel lines,
the alternate interior angles are equal
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Appendix (cont.)
Step 4. Label the diameter of the pipe d, and
the length to cut off from the top as ltop.
Then, we see the relationship, tan ? d
/ ltop
So, ltop d / tan ?
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Appendix (cont.)
Step 5. Similarly, near the bottom, label
again the diameter d, and the length of the piece
to cut off, lbottom.
Step 6. We can find the angle complement to ?,
which is 90 - ?.
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Appendix (cont.)
Step 7. To find the length lbottom,
simply calculate, tan(90 - ?) d /
lbottom
Or, lbottom dtan(90 - ?)
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Appendix (cont.)
Thus, you have reviewed enough trigonometry to
see how to calculate the lengths to trim.
The calculations for trimming the top and bottom
pieces are done in the same fashion, just with a
different angle ?.
If you linked to this appendix from the slide
show, click here to return back to the
presentation.
End of module.