| Optimization: Polygon of Constraints |
DECISION MAKING OR OPTIMIZATION:
When we have a situation with more than one inequality to solve simultaneously,
and/or we have to maximize or minimize a given function known as a target objective,
we construct a polygon of constraints.
POLYGON OF CONSTRAINTS :
When we graph a system of inequalities, we find a region of the plane containing all points that satisfy all the inequalities in the system. This is known as the polygon of constraints.
The coordinates of all the points in this polygon are the solution set.
If we need to maximize or minimize a target objective -- such as monthly profit or weekly costs -- we substitute the x and y values at the vertices of the polygon of constraints into the inequality statements to verify which pair of coordinates creates a maximum or minimum as required.
Example 1 :
The student council of a school is planning a trip. It has a maximum of $600 available to pay for transportation. Regular buses which seat 40 people cost $80 each to rent and minibuses which seat 20 people cost $60 each to rent. A minimum of 240 students is required for the trip to take place. Give 3 possibilities for renting regular buses x and minibuses y.
Solution :
| the inequality for the number of people is | 40x + 20y m 240 |
| the inequality for the cost of renting buses is | 80x + 60y [ 600 |
| dividing both inequalities by 20 | 2x + y m 12 t y m -2x + 12 |
| 4x + 3y [ 30 t y [ (-4/3) x + 10 |
When we graph these lines and shade the area which satisfies the inequality, we get:
Since we can't rent a part of a bus, any point in the polygon of constraints
with integer coordinates, is a solution. Three possibilities for renting buses are:
(6, 0) , (3, 6) or (4, 4)
When dealing with a real situation such as the one in the example above, we must be careful not to include parts of the coordinate plane which don't suit reality. For example, if the x-axis represents time or number of buses (as in the example), we cannot include negative values since neither time nor number of buses can be negative. In most real situations, we will use only the first quadrant or the first and fourth quadrants.
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Translating Words into Math
The hardest part of solving a question on optimization is translating the words of the problem into mathematical statements. The key phrases "at least" and "at most" are generally the source of the confusion -- but sometimes the convoluted manner in which the question is presented doesn't help either.
at least is a floor. Everything is on or above it so at least is equivalent to m .
at most is a ceiling. Everything is on or below it so at most is equivalent to [ .
Example 2: Write linear inequalities to represent these situations:
a) The perimeter of a rectangle is at least 80 cm.
b) A toy store owner bought at most $350.00 worth of blue balls that cost 55¢ each and red balls that cost 67¢ each.
c) Harry buys a bouquet of purple flowers at $1.25 each and yellow flowers at $1.10 each for his mom. He can spend no more than $22 for the bouquet and he wants no less than 8 purple flowers in the arrangement.
Solutions:
a) 2(l + w) m 80
b) 0.55b + 0.67r [ 350
c) 1.25p + 1.10y [ 22, and p m 8
It should be obvious now that no less than is the same as at least, and
no more than is the same as at most.
Note: when listing the inequalities in the system, we must indicate that the variables are all positive therefore we should see x ¥ 0, y ¥ 0 in each question.
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The Target Objective
The target objective is a function that we must optimize -- find a minimum or maximum.
It is a statement of equality, not an inequality. ex: P = 5x + 2 y
It could represent the profit on t-shirt sales or the number of tickets bought.
We deal with this function once we've set up the polygon of constraints. Here's an example.
Example 3:
A Bippy maker produces dotted and striped Bippies. The striped ones are twice as popular as the dotted ones so he makes twice as many striped as dotted Bippies per week to satisfy the demand. He can make at most 30 Bippies per day. The profit on a dotted Bippy is $2.15 whereas the profit on a striped Bippy is $2.75. Under these conditions, what is the maximum daily profit for our Bippy maker?
if we let x = # of dotted Bippies made in a day
and y = # of striped Bippies made in a day then
the target objective here is P = 2.15x + 2.75y
The inequalities are: x + y [ 30 , and y m 2x
The point of intersection is (10, 20). It's obvious that (0, 0) will give a minimum.
We substitute (0, 30) and (10, 20) into the target objective P = 2.15x + 2.75y.
At (0, 30), we get 2.15(0) + 2.75(30) = $82.50
At (10, 20), we get 2.15(10) + 2.75(20) = $76.50
so our Bippy maker should make 30 striped and 0 dotted Bippies per day to
maximize his daily profit. His maximum daily profit will be $82.50.
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Hint: if you're not sure what the inequality is, write the two variables with a blank between them then figure out which is larger and put in the correct inequality symbol.
example: ... she wants to have at least twice as much land planted in carrots as in beans (this is the way they talk in these problems!)
If c = area of land planted in carrots and b = area of land planted in beans
Since she wants more carrot-land than bean-land we know that c > b so b will be multiplied by 2 to reach equality. The statement becomes: c m 2b.
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Example 4:
Jessica raises money by painting designs on vases and sugar bowls which she sells at a marché aux puces. It takes 2 hours to paint a vase and 3 hours to paint a sugar bowl. She can devote a maximum of 120 hours per school semester to her project during which she expects to paint a maximum of 50 items. She must paint at least 10 sugar bowls to satisfy customer demand.
If a vase sells for $12 and a sugar bowl sells for $8,
how many of each must she sell to maximize her profit?
Let x = # of vases, and y = # of sugar bowls
The inequality about her time is: 2x + 3y [ 120
The inequalities about the number of items painted is: x + y [ 50, and y m 10
The target objective or function to maximize is her profit P = 12x + 8y
2x + 3y [ 120 t y [ -2/3x + 40, and x + y [ 50 t y [ - x + 50
The vertices of the polygon of contraints for this situation are:
(0, 40), (30, 20)(the point of intersection), and (40, 10) -- where -x + 50 meets y = 10.
When we substitute the coordinates of these vertices into P, we find that (40, 10) yields a maximum profit of $560 so Jessica should paint 40 vases and 10 sugar bowls
for a maximum profit of 12(40) + 8 (10) = $560.00.
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Practice
1) A fisherman who sells lobsters at $8.70 each and crabs at $9.60 each says that on an average day he catches no less than 35 crabs but no more than 60. His catch usually has at most twice as many lobsters as crabs. He also claims he's never caught more than 140 shellfish in a day.
a) Write the system of inequalities to represent this situation.
b) Write the target objective P to represent his daily profit.
c) Draw a polygon of constaints and determine his maximum daily revenue under these restrictions.
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2) Note: this is the toughest question on this topic I've ever seen on a final exam. It involves using a "dummy" variable which makes things a little less clear.
A campground with 150 camp sites rents both tent and trailer sites. A trailer site costs twice what a tent site does for a day. Tent site users generally consume an average of 10 litres of water per day while trailer site users consume an average of 30 litres of water per day. The campground has access to no more than 3300 litres of water each day. The maximum daily profit for the campground is $4800.
How many of each type of site (tent or trailer) are there in this campground?
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Solutions
1) a) Let x = # of lobsters caught daily
b) P = 8.70 x + 9.60 y
c)
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2) Let x = # of tent sites in the campground
Let y = # of trailer sites in the campground
Let a = daily cost of renting a tent site,
therefore 2a = cost of trailer site,
therefore ax + 2ay = $4800 (max profit when all sites rented)
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