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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 21 "Optimiza
tion Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" 
{TEXT -1 14 "Daniele Arcara" }}{PARA 257 "" 0 "" {TEXT -1 25 "Departme
nt of Mathematics" }}{PARA 257 "" 0 "" {TEXT -1 21 "University of Geor
gia" }}{PARA 257 "" 0 "" {TEXT -1 18 "Copyright (c) 2002" }}}{SECT 0 
{PARA 3 "" 0 "" {TEXT -1 21 "Optimization Problems" }}{PARA 0 "" 0 "" 
{TEXT -1 502 "One of the main application of Calculus is in solving th
e so-called \"Optimization Problems\".  These are problems in which we
 try to find out the best way to maximize or minimize a certain quanti
ty.  For example, we might want to minimize the cost of some operation
, or we might want to maximize the volume of a box that we are trying \+
to construct with the material we have.  Or maybe our company wants to
 maximize the profit it could make in buying certain material and sell
ing back a finished product." }}{PARA 0 "" 0 "" {TEXT -1 254 "In your \+
Calculus class, you will develop certain tools that will help you in s
olving these problems in a precise way, but we shall try to get an ide
a of what the answer should/would be in this lab by looking at the gra
ph of certain appropriate functions." }}}{SECT 0 {PARA 3 "" 0 "" 
{TEXT -1 30 "Example 1: Area of a rectangle" }}{PARA 0 "" 0 "" {TEXT 
-1 197 "Suppose that want to build a rectangular garden that contains \+
600 feet squared and that you want to put a fence around it.  How shou
ld you do this in order to minimize the amount of fencing to use?" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 180 "Let us start with drawing the pic
ture of a rectangle, and labeling the sides.  Since you do not know ho
w to do this with Maple, draw it on a piece of paper.  If we call the \+
length " }{TEXT 264 1 "l" }{TEXT -1 16 " and the height " }{TEXT 263 
1 "h" }{TEXT -1 79 ", the amount of fencing that we use is given by th
e perimeter of the rectangle:" }}{PARA 256 "" 0 "" {TEXT -1 14 "Perime
ter = 2 " }{TEXT 266 1 "l" }{TEXT -1 5 " + 2 " }{TEXT 265 1 "h" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 249 "We cannot go ahead and \+
graph this function, since it is a function of two variables.  What to
 do then?  Well, we are going to use the fact that we want our garden \+
to contain 600 feet squared, which means, we want the area of the rect
angle to be 600." }}{PARA 0 "" 0 "" {TEXT -1 41 "Since the area of a r
ectangle is given by" }{TEXT 267 0 "" }}{PARA 256 "" 0 "" {TEXT -1 7 "
Area = " }{TEXT 268 3 "l h" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT 
-1 37 "we can set it equal to 600, and find " }{TEXT 270 1 "h" }{TEXT 
-1 18 " as a function of " }{TEXT 269 1 "l" }{TEXT -1 12 " as follows:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve(l*h=600,h);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 188 "Now we know that, in order to hav
e the area equal to 600, we need to choose our height to be 600 divide
d by the length.  This implies that we can write our perimeter as a fu
nction of just " }{TEXT 272 2 "l " }{TEXT -1 16 "by substituting " }
{TEXT 273 1 "h" }{TEXT -1 9 " = 600 / " }{TEXT 271 1 "l" }{TEXT -1 1 "
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "subs(h=600/l,2*l+2*h);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Let us call " }{TEXT 274 1 "P
" }{TEXT -1 41 " the perimeter function as a function of " }{TEXT 276 
1 "l" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "P:=
l->2*l+1200/l;" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 "
Now we can plot the graph of the perimeter, and try to guess which len
gth would minimize it." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "p
lot(P(l),l=0..100,y=0..1000);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "
It seems that the minimum is between 20 and 30 feet: let us zoom in." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "plot(P(l),l=20..30);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 215 "To minimize the perimeter, we sho
uld pick the length to be about 24.5 feet.  To find the corresponding \+
height, we are going to plug this value back into the relation we foun
d above between the height and the length." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 19 "subs(l=24.5,600/l);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 227 "It seems that a square with a side of 24.5 feet would mi
nimize the amount of fencing that we need to enclose 600 feet squared \+
of surface.  This is actually true, as you shall see in your Calculus \+
class later on in the semester." }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 
16 "Project - Part 1" }}{PARA 0 "" 0 "" {TEXT -1 229 "Suppose now that
 you already have the fencing, and you have 120 feet of it.  What dime
nsions would maximize the area that you can enclose with your fencing \+
using a rectangular shape?  Does a square still give you the best answ
er?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 21 "Example 2: \+
A soda can" }}{PARA 0 "" 0 "" {TEXT -1 325 "Consider the following sit
uation: We want to make a soda can (in aluminium) that contains 1/2 li
ter of liquid (about 17 fluid ounces) to sell our favorite drink.  Ass
uming that the thickness of the material is uniform, we want to find t
he dimensions of the can that would minimize the amount of aluminium t
hat we have to use." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 250 "To answer \+
this question, let us start with drawing a picture of the soda can.  A
gain, since you cannot do it with Maple, use a piece of paper.  For us
, it will look like a cylinder.  Draw a cylinder, and label the quanti
ties that determine it.  Call " }{TEXT 257 1 "r" }{TEXT -1 55 " the ra
dius of the base (and top) of the cylinder, and " }{TEXT 258 1 "h" }
{TEXT -1 299 " the height of the soda can.  We said that we want to mi
nimize the amount of material used, and this means that we want to min
imize the surface area of the can.  The surface area is given by the a
rea of the two circles (the top and the bottom of the can), and by the
 curved surface area on the side." }}{PARA 256 "" 0 "" {TEXT -1 21 "To
tal Surface Area = " }{XPPEDIT 18 0 "2*Pi*r^2+2*Pi*r*h;" "6#,&*(\"\"#
\"\"\"%#PiGF&%\"rGF%F&**F%F&F'F&F(F&%\"hGF&F&" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 147 "Unfortunately for us, we cannot go ahead
 and draw the graph of this function to try to find the minimum, becau
se there are two variables involved: " }{TEXT 259 1 "r" }{TEXT -1 5 " \+
and " }{TEXT 260 1 "h" }{TEXT -1 294 ".  But this is what we are going
 to do: we know that we want only 17 fluid ounces in our can, and so t
he volume (in inches cubed) of the can should be 17 multiplied by 1.80
469 (since there are approximately 1.80469 inches cubed in a fluid oun
ce).  Since we know that the volume of a cylinder is" }}{PARA 256 "" 
0 "" {TEXT -1 9 "Volume = " }{XPPEDIT 18 0 "Pi*r^2*h;" "6#*(%#PiG\"\"
\"*$%\"rG\"\"#F%%\"hGF%" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 
20 "we can conclude that" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "Pi*r^2*h
" "6#*(%#PiG\"\"\"*$%\"rG\"\"#F%%\"hGF%" }{TEXT -1 16 " = 17 * 1.80469
." }}{PARA 0 "" 0 "" {TEXT -1 61 "With the help of Maple, we can now u
se this equation to find " }{TEXT 261 1 "h" }{TEXT -1 18 " as a functi
on of " }{TEXT 262 1 "r" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "solve(Pi*r^2*h=17*1.80469,h);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 214 "Now that we know that, in order to have 17 fluid ou
nces in our can, we need the height of the can to be approximately equ
al to 9.76566 divided by the radius of the can squared, we can define \+
our total surface area " }{TEXT 275 1 "A" }{TEXT -1 29 " as a function
 of the radius:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "subs(h=9
.765661364*1/(r^2),2*Pi*r^2+2*Pi*r*h);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "A:=r->2*Pi*r^2+19.53132273*Pi/r;" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 193 "And now we can plot the graph of the surface area
 to see where the minimum should/would be.  Note that we are only goin
g to use a positive domain since the radius of the can cannot be negat
ive." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot(A(r),r=0..10,y
=0..1000);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "It seems that the m
inimum is when the radius is a little less than 2; let us zoom in:" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(A(r),r=1..3);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 280 "Now it is more clear that the rad
ius should be about 1.7 inches.  To finish the problem, let us find th
e corresponding height of the can.  Remember that we already have the \+
height expressed as a function of the radius, so we can just substitut
e our values for the radius in there:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "subs(r=1.7,9.765661364*1/(r^2));" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 91 "The height should be about 3.38 inches, and the to
tal surface area is 54.25 inches squared." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 7 "A(1.7);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
14 "evalf(A(1.7));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{SECT 0 {PARA 3 "" 0 "" {TEXT -1 16 "Project - Part 2" }}{PARA 0 "" 0 
"" {TEXT -1 343 "Suppose that you have the opposite problem now: you h
ave a total amount of 60 inches squared of aluminium, and you want to \+
make a soda can out of it.  You want the can to be able to contain the
 maximum amount of liquid possible.  Find the dimensions of the can th
at would do that.  What is the total amound of fluid that would fit in
 your can?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 16 "Project - P
art 3" }}{PARA 0 "" 0 "" {TEXT -1 483 "Your boss comes to you one day,
 and gives you a bunch of pieces of cardboard, telling you to make box
es out of them.  The pieces of cardboard are actually squares, and eac
h side is 20 inches long.  You want to impress your boss, and you are \+
going to make the boxes that contain the largest amount of \"stuff\" i
n them.  Which means, you want to cut off the corners, fold the sides \+
up, and get a box of the largest possible volume.  What is the volume \+
of the boxes that you constructed?" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 3 "" 0 "
" {TEXT -1 16 "Project - Part 4" }}{PARA 0 "" 0 "" {TEXT -1 120 "Selec
t two or three problems out of your Calculus book from the Section on \+
Optimization, and solve them as we did above." }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 4 0" 21 }
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