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Math 108, Lab 5, Optimization In mathematics, the term “optimization” means “making the best choice of something”. What “best” means will deptnd on the application. It might mean the lowest cost, or the highest profit, or the largest area, or.... The hard part of optimization is usually finding a formula ^ that describes the desired output (or target, goal, or objective variable ) in terms of a variable that controls that outcome, where is often subject to constraints imposed by the application. Finding the formula for the goal variable is ours to do. Maple can’t help us until we have it.
1. Planning an athletic field. We want to build an athletic field consisting of a central rectangle with two semicircular areas attached to opposite sides of the rectangle. The perimeter of the field will be a 440 yard track. We want the central rectangular area to be as large as possible. What should we do?
x yards
y yards
^ Our target variable is the area of the central rectangle. Area of a rectangle is the product of its side lengths, ^ and^ in the picture above. So it appears that our goal variable^ ^ depends on two variables^ and^ . But and cannot be just anything at all – they must be chosen so that the perimeter of the field is 440 yards and that constraint allows us to solve for one of and in terms of the other and express in terms of one single variable. The perimeter of the field consists of two straight stretches, each units long, plus to semi-circular arcs. If they were put together, those two semicircular arcs would yield a full circle whose circumference is . So is another variable that we must consider? No, because we see from our picture above that ^ . Therefore, the length of the two semi-circles is ^ ^ ^ ^ . Adding in the lengths of the two straight segments gives the equation !^ "$#%#^ & so that ')(*(,+.-0/,132 4. Putting together what we have figured out, we have
:^ 65^ 7^ ^98 <;= > ? 65 A@<B9CD@>E 'FE#%#^ &G^ H I^ KJL ? M8^ [Remember to use a capital P when writing the number Pi.] Because the entire perimeter of the field must be 440 yards, the length of cannot exceed 220 yards, and clearly ON$&^. So, we PRQESUTDE: (^) <;= > ? VJ 7& WXW (^) % &% M
and we see that there is a maximum value for area somewhere between 100 and 120. To find out exactly where, we useY < ?Z[5
Y
Z]E:^ <;= > ? VJ M
Z
Y
Q^A]@SQ^ [0
Y
< Z3[ _&^ `8^ [which gives 110]. To find out the area of the largest possible rectangular field, we
@<B CM@>E Z
Y
(^) Q^ (^) J: <;= M8 (^) [which gives /(,/++ 4 ]
and then <[0 Q ^ M8^ [which gives 7703.099244]. To find the other dimension of the field, we order <[0 Q @<B^ CM@>E^ Z
Y
Q^ J "FE#%# &G H . I M
and that gives & W^ & ?#^ #^. In this problem, the and variables are measured in yards, while is in square yards.
2. Minimizing inventory costs. When a company purchases and stores products for sale (= the inventory), it often borrows the required money from a bank, planning to pay back the loan with interest after the goods are sold. That interest is part of the cost of doing business. Furthermore, companies might also face costs for keeping inventories in storage. Together these kinds of costs are often called carrying costs. By making frequent orders for small amounts of goods, companies can keep these costs low. Unfortunately, the process of placing orders also costs money, in terms of employee time, for example, and the more frequently one orders, the higher will be the company’s ordering costs. Thus, the very same business practices that decrease carrying costs will tend to increase ordering costs. The problem is to order just frequently enough so that the sum TST^ QMS?@<T^ > Z: DS?@^ T S
Y
(^) < Z : MS@ T is as small as possible. A bookstore anticipates selling 1,000 copies of an expensive reference book each year, at a steady rate spread over the entire year. Consequently, the store decides to place a sequence of equally spaced, equal-sized orders with the publisher during the year. The store borrows the money to buy the merchandise, and ends up paying $ in interest per year for each book purchased. Ordering costs are estimated to be $50 each time an order is placed. How many times per year should the store order, and how many copies of the book should it order each time if the goal is to minimize the total inventory cost for the book? The solution of this problem depends to some degree on how the bank charges interest on loans (and there are many ways). We will assume that the company decides to borrow all of the money needed for a single order at the beginning of the year, and pay interest on it all year long at the rate of $6 per book per year. Let : denote the number of times per year that we place orders. Then S
Y
< MS?@<T^5 &^ :^8
and each time we place an order, the number of books we must order is S
Y
< @Z% 5 &%&&^ ?:^8
Given our decision about borrowing money (see above), our interest carrying costs will be Z: DS?@^ T 5 S
Y
< @Z 8
so that our inventory costs will be DS?@ T 5 (^) > Z: DS?@ T S
Y
< DS?@ T`
All we know about the number : is that it cannot be less than 1 and probably cannot be more than 365. Therefore, to look at the MS@^ T graph, we PRQESUTD DS?@ TJ: WW % M
That graph makes it clear that a minimum cost occurs somewhere below : &^. To get more detail, we look at
Notice that the formula displayed for DS?@^ T contains the letters J : J , and @, and from MAPLE’s point of view, they all look like the same kind of thing. Of course, we know that : is the only real variable here and that the other letters stand for constants. The existence of all of those letters has at least one unfortunate consequence: MAPLE cannot graph the MS@^ T formula for us.
[Try writing PRQESTD MS@ TJ: WXW (^) &%&% M and you will get an empty plot error message.]
Therefore, we must proceed without knowing whether there is a maximum or minimum MS?@<T^ value for us to find. Our first step isY < ?Z[5
Y
Z]^ DS?@^ TJ: `
[Footnote: Given that there are so many letter symbols in the MS@^ T formula, it is not surprising that we must tell MAPLE which one is the real variable of the problem when we order MAPLE to find the derivative. Now you see why we have been writing the in
Y
Z]\^ J all along.] Given
Y
< Z3[ , we next want to solve
Y
< Z3[ &. Because this is an algebra situation rather than a decimal number situation, we use @SQ^ [ rather than ]@SQ^ [ and we write Z
Y
Q:^5 A@S?Q^ [0
Y
< Z3[ _&^ J:] M
which gives
Z
Y
Q:^ ^
=^
W
Because of its meaning in this problem, we know that Z
Y
Q:^ is positive, so we choose Z
Y
Q:^5 / / ^11
(^).
[If you want, try ]@S?Q^ [0
Y
< Z3[ _&^ J:] M8^. You will get an error message.]
At this point we have a candidate value of MS@ T : for getting a maximum, minimum, or plateau point on the graph of
. But we can’t graph the MS@^ T function without knowing the values of "J , and @, so how can we tell whether we have a minimum, a maximim, or neither? There is a way. Look atY < ?Z[ !5
Y
Z]^
Y
< ?Z[RJ : `
and you see that the second derivative of cost is 1 ^. Whatever the values of J : and @ might be in our application,
they certainly will be positive so that we can be sure that the second derivative of the cost function will be positive, Now we recall the second derivative test, and we conclude that the value Z
Y
Q:^ where
Y
< Z3[ & will actually give us a cost minimum, as desired. The final step is to determine the ideal order size, now that we know that the ideal number of times per year to order is Z
Y
Q:^. We will divide the total quantity in Z
Y
QE:^ equal orders, so we get Z
Y
QES^
Y
< @Z 5 Z
Y
Q:^5
and MAPLE reports Z
Y
QES^
Y
< @Z 1 1 /
1 . Squaring both sides shows that to be the same as /,1^ ^1 , the
formula given in books on inventory control.
Homework, M108 Lab 5 As usual, you may work in groups of size up to three. Groups should hand in a single solution sheet with all group members’ names on it. This homework is due when your lab meets next week.
- This problem is somewhat related to Example 1, above. We are trying to build a window consisteing of a rectangle with a semi-circle on top, as in figure 1. The total area of the window is constrained to be 1000 square inches, and we want our window to have the smallest possible outside perimeter. What should be the dimensions of the window?
h inches
w inches
total window area = 1000 sq in
a) Let be the vertical height of the rectangular part of the window and let ; be the width. The outside perimeter consists of two lengths of units each, plus one length of ; units, plus a semicircular component. Using ; as the variable, give a formula for the length of that semicircular component. b) Now give a formula for the total length of the window’s external perimeter in terms of the variables ; and . (c) Write a constraint equation that relates the variables ; and . This will involve adding together the area of a rectangle and of a semi-circle to get 1000. This is NOT a MAPLE problem. d) Now solve that constraint equation for in terms of ; , either by hand or using MAPLE. Watch out for signs. e) Use your results from b) and d) to write the formula for perimeter in terms of the variable ;. This is not necessarily a MAPLE problem. f) What value of ; gives the least possible perimeter for the window? g) What is the least possible perimeter for the window? [near 120 feet] (h) How do you know that you found the least possible perimeter, rather than the greatest possible perimeter?
