Example
A decorative item dealer deals in only two items - wall hangings and
artificial plants. He has 15,000 dollars to invest and a space to store all the
most 80 pieces. A wall hanging costs him 300 dollars and an artificial plant
150 dollars. He can sell a wall hanging at a profit of 50 dollars and an
artificial plant at a profit of 18 dollars. Assume that he can sell all the
items that he buys. Let us make a mathematical model to maximize his profit
with the given conditions.
Sol.
Let the person sells x number of wall hangings and y number of artificial
plants.
We tabulate the given data as follows:
Since the object in this
problem is to maximize the profit, let
Z = 50x + 18y be the profit function.
Since the dealer has space availability of (to store) 80 pieces
Since
he can invest 15,000 dollars, his cost cannot be more than 15,000
Therefore
300x + 150 y ? 15,000 dollars
The mathematical model for this problem is
Maximize Z = 50x +
18y
(objective function)
Subject to the condition
with non-negativity conditions
Note that in both the examples the conditions are linear inequalities; these mathematical
models which tells to optimize (minimize or maximize) the objective function Z
subject to certain condition on the variables is called a Linear programming
problem (LPP).
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Q1/ Find two positive numbers whose sum is at least 15 and whose
difference is at the most 7 such that their product is maximum
Before solving this problem, we have to state the problem
Step 1:
We have to choose two positive numbers. Let the two positive numbers be x
and y. This x and y are decision variables.
Step 2:
Our objective is to maximize the product xy
Let Z = xy. We have to maximize Z
Step 3
We have the following conditions on the variables x and y.
a) Sum of the numbers is at least 15
i.e., x + y 15
b) Difference of the numbers is at most 7
i.e., x - y 7
c) Since the x and y should be positive, we have two more conditions
x > 0, y > 0 as Linear
constraints
The mathematical model for this problem is maximize the objective function Z =
xy
Subject to:
x + y 15
x - y 7
x > 0, y > 0
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A Diet Problem
Suppose the only foods available in your local store are potatoes and steak.
The decision about how much of each food to buy is to made entirely on dietary
and economic considerations. We have the nutritional and cost information in
the following table:
:
Per unit
of potatoes Per unit
of steak Minimum
requirements
Units of carbohydrates 3 1 8
Units of vitamins 4 3 19
Units of proteins 1 3 7
Unit cost 25 50
The problem is to find a diet (a choice of the numbers of units of the two
foods) that meets all minimum nutritional requirements at minimal cost.
a. Formulate
the problem in terms of linear inequalities and an objective function.
b. Solve the problem geometrically.
c. Explain how the 2:1 cost ratio (steak to potatoes) dictates that the
solution must be where you said it is.
d. Find a cost ratio that would move the optimal solution to a different
choice of numbers of food units, but that would still require buying both steak
and potatoes.
e. Find a cost ratio that would dictate buying only one of the two foods
in order to minimize cost.
a) We begin by setting the constraints for the problem. The first constraint
represents the minimum requirement for carbohydrates, which is 8 units per some
unknown amount of time. 3 units can be consumed per unit of potatoes and 1 unit
can be consumed per unit of steak. The second constraint represents the minimum
requirement for vitamins, which is 19 units. 4 units can be consumed per unit
of potatoes and 3 units can be consumed per unit of steak. The third constraint
represents the minimum requirement for proteins, which is 7 units. 1 unit can
be consumed per unit of potatoes and 3 units can be consumed per unit of steak.
The fourth and fifth constraints represent the fact that all feasible solutions
must be nonnegative because we can t buy negative quantities.
constraints:
3X1 + X2 ? 8,
4X1+ 3X2 ? 19,
X1+ 3X2 ? 7,
X1? 0, X2 ? 0
Next we plot the solution set of the inequalities to produce a feasible region
of possibilities.
c) The 2:1 cost ratio of steak to potatoes dictates that the solution must be
here since, as a whole, we can see that one unit of steak is slightly less
nutritious than one unit of potatoes. Plus, in the one category where steak
beats potatoes in healthiness (proteins), only 7 total units are necessary.
Thus it is easier to fulfill these units without buying a significant amout of
steak. Since steak is more expensive, buying more potatoes to fulfill these
nutritional requirements is more logical.
d) Now we choose a new cost ratio that will move the optimal solution to a
different choice of numbers of food units. Both steak and potatoes will still
be purchased, but a different solution will be found. Let s try a 5:2 cost
ratio.
d) Now we choose a new cost ratio that will move the optimal solution to a
different choice of numbers of food units. Both steak and potatoes will still
be purchased, but a different solution will be found. Let s try a 5:2 cost
ratio.
d) Now we choose a new cost ratio that will move the optimal solution to a
different choice of numbers of food units. Both steak and potatoes will still
be purchased, but a different solution will be found. Let s try a 5:2 cost
ratio.
Thus, the optimal solution for this cost ratio is buying 8 steaks and no
potatoes per unit time to meet the minimum nutritional requirements.
________________________________________
A Blending Problem
Bryant s Pizza, Inc. is a producer of frozen pizza products. The company makes
a net income of $1.00 for each regular pizza and $1.50 for each deluxe pizza
produced. The firm currently has 150 pounds of dough mix and 50 pounds of topping mix.
Each regular pizza uses 1
pound of dough mix and 4 ounces (16 ounces= 1 pound) of topping mix.
Each deluxe pizza uses 1
pound of dough mix and 8 ounces of topping mix.
Based on the past demand per week, Bryant can sell at least 50 regular pizzas
and at least 25 deluxe pizzas. The problem is to determine the number of
regular and deluxe pizzas the company should make to maximize net income.
Formulate this problem as an LP problem.
Let X1 and X2 be the number of regular and deluxe pizza, then the LP
formulation is:
Maximize X1 + 1.5 X2
Subject to:
X1 + X2 ? 150
0.25 X1 + 0.5 X2 ? 50
X1 ? 50
X2 ? 25
X1 ? 0, X2 ? 0
EXAMPLES
OF LINEAR PROGRAMS
Red
or White?
A Juice maker would like to decide how many bottles of red juice and how
many bottles of white juice to produce. Given his expertise is in
red making he can sell a bottle of red for $12 while he can only sell a bottle
of white for $7. Clearly the maker would seek to maximize his profits, and,
having recently
completed a course in mathematical modeling, proceeds to construct the
objective function
F(x1; x2) = 12x1 + 7x2
where the decision variables are the number of bottles of red to produce x1 and
the number of bottles of white to produce, i.e., x2. Aging milk in wooden or
glass-lined vats is an integral component of the
production process, but due to limited space, the juice must be aged for a
limited time. The maker has determined that red should be aged two years per
bottle and white one year per bottle and his facilities allow that each batch
is limited to 10,000 bottle-years (5 bottles of red and 3 bottles of white
require a total
of 13 bottle years ripening time). Thus the maker formulates a constraint:
2x1 + x2 <= 10000
Also the volume of grapes that may be processed is limited and it takes 3 gallons of grapes to
make a bottle of red and two gallons of grapes to make a bottle of white.
Furthermore, the winery can only process a total of 16,000 gallons of
grapes for each batch. Thus, the maker produces the additional constraint
3x1 + 2x2 <= 16000
Now the maker would like to determine how many bottles of each type to produce
as well as how much money he will expect to make. Note that we must also
require that negative bottles are not allowed so
x1 >= 0 and x2 >= 0
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Find the altitude to the base of an isosceles triangle with a side length of 4,
such that the area of the triangle is maximal.
A basic type of problems in calculus is the minimization problem.
This solution deals with both minimization process of an area of an isosceles
triangle
• Constructing
an Altitude - Construct an equilateral triangle. Then construct one of its
altitudes. If the sides each have length 1, how long is the altitude -
Construct an equilateral triangle. Then construct one of its altitudes. If the
sides each have length 1, how long is the altitude
• Geometric mean - Prove that the length of the altitude to the hypotenuse
of a right triangle is the geometric mean of the lengths of the segments into
which the altitude partians the hypotenuse.
• Proofs - Triangles - 1- if we have a triangle ABC, then prove that the
internal and external bisectors of the angle of a triangle are perpendicular
(assume for angle A) 2- prove that given triangle ABC with the altitud ...
• Size of Triangles - 28. Find the area and perimeter of an equilateral
triangle whose altitude measures 6
cm. 29. If the area of an equilateral triangle is 100 sq
cm, find the length of a side.
• Geometry/construction - -construct triangle ABC given the length of one
side BC, the length of the median to BC and the length of the altitude from B