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```Week
4
Linear Programming
Junior: Nguyen Anh Cuong
Exercise 1
A nutritionist is planning a menu consisting of two main foods A and B. Each ounce
of A contains 2 units of fat, 1 unit of carbohydrate, and 4 units of protein. Each ounce of
B contains 3 units of fat, 3 units of carbohydrates, and 3 units of protein. The nutritionist
wants the meal to provide at least 18 units of fat, at least 12 units of carbohydrate, and
at least 24 units of protein. If an ounce of A costs 20 cents and an ounce of B costs 25
cents, how many ounces of each food should be served to minimize the cost of the meal
yet satisfy the nutritionist’s requirements?
Solution
Let x and y denote the number of ounces of foods A and B, respectively, that are served.
The number of units of fat contained in the meal is 2x + 3y, so that x and y have to
satisfy the inequality
2x + 3y ≥ 18.
Similarly, to meet the nutritionist’s requirements for carbohydrate and protein, we must
have x and y satisfy
x + 3y ≥ 12 and 4x + 3y &gt; 24
Of course, we also require that x &gt; 0 and y &gt; 0. The cost of the meal, which is to be
minimized, is
z = 20x + 25y
Thus, our mathematical model is:
Find values of x and y that will minimize z = 20x + 25y subject to the restrictions
2x + 3y ≥ 18
x + 3y ≥ 12
4x + 3y ≥ 24
x ≥ 0,
y ≥ 0.
Follow image above, we have that z will be minimize equal 160 if and only if x = 3, y = 4.
Exercise 2
An appliance company manufactures heaters and air conditioners. The production
of one heater requires 2 hours in the parts division of the company and 1 hour in the
assembly division of the company; the production of one air conditioner requires 1 hour
in the parts division of the company and 2 hours in the assembly division of the company.
The parts division is operated for at most 8 hours per day and the assembly division is
operated for at most 10 hours per day. If the profit realized upon sale is 30 per heater
and 50 per air conditioner, how many heaters and air conditioners should the company
manufacture per day so as to maximize profits?
Solution
Let the number of heaters that can be produced in a day is x and The number of air
conditioner that can be produced in a day is y.(x, y ≥ 0)
Following the topic, we easy to find inequality:
The parts division is operated for at most 8 hours per day then satify that 2x + y ≤ 8.
The assembly division is operated for at most 10 hours per day then satify that x+2y ≤ 10.
When we denoted that maximize profits is z = 30x + 50y so we will find x, y sauch that
z is maximum value.
2x + y ≤ 8
x + 2y ≤ 10
x, y ≥ 0
Follow image above, we have that z will be maximum equal 260 if and only if x = 2,
y = 4.
Exercise 3
An oil company owns two refineries, say refinery A and refinery B. Refinery A is
capable of producing 20 barrels of gasoline and 25 barrels offuel oil per day; refinery B is
capable of producing 40 barrels of gasoline and 20 barrels of fuel oil per day. The company
requires at least 1000 barrels of gasoline and at least 800 barrels of fuel oil. If it costs 300
per day to operate refinery A and 500 per day to operate refinery B, how many days
should each refinery be operated by the company so as to minimize costs?
Solution
Let x be of number days for refinery A and y be number of days for refinery B.
then the quantity to be minimized, namely cost, is given by z = 300x + 500y. Note
that the company must incur some positive cost since it is constrained by the minimum
petroleum requirements. Refinery A is capable of producing 20 barrels of gasoline per day
and refinery B is capable of producing 40 barrels of gasoline per day. Hence, the total
amount of gasoline produced is 20x + 40y barrels. Inasmuch as at least 1000 barrels of
gasoline is required by the company, we have the constraint
20x + 40y ≥ 100
Similar reasoning applied to the fuel oil yields the constraint
25x + 20y ≥ 800
Satify that x ≥ 0, y ≥ 0.
So following the image we conclude that z be minimum if and only if z = 13500 with
respect to x = 20 and y = 15.
Exercise 4
A manufacturer of artificial sweetener blends 14 kilograms of saccharin and 18 kilograms of dextrose to prepare two new products: SWEET and LO-SUGAR. Each kilogram
of SWEET contains 0.4 kilogram of dextrose and 0.2 kilogram of saccharin, while each
kilogram of LO-SUGAR contains 0.3 kilogram of dextrose and 0.4 kilogram of saccharin.
If the profit on each kilogram of SWEET is 20 cents and the profit on each kilogram of
LO-SUGAR is 30 cents, how many kilograms of each product should be made to maximize
the profit?
Solution
Let number of kg SWEET is x and number of LO-SUGAR is y satify that x, y ≥ 0.
So Dextrose have all 18 kilogram then we have 0, 4x + 0, 3y = 18 and Saccharin have all
14 kilogram then we have 0, 2x + 0, 4y = 14.
Therefore, Let z = 30x + 30y be maximize the profit so we have using graph menthod to
solve this problem.
Easy to see that D is the point such that z = 30x + 30y be maximize with x = 30 and
y = 20.
Exercise 5
Let x2 = x4 − x5 with respect to x4 , x5 ≥ 0.
So we have standart form:
−3x1 + 4x3 − x4 + x5 → min
Such that


−7x1 + 2x4 − 2x5 + 3x3 ≥ 4






 2x1 − 4x4 + 4x5 − 8x3 ≥ 3
D : −2x1 + 4x4 − 4x5 + 8x3 ≥ −3




−5x1 + 3x4 − 3x5 + 2x3 ≥ −9




x
≥0
1,3,4,5
Exercise 6
Let
x1 = x4 − 2
x2 = x5 − x6
x3 = x7 − x8
such that x4,5,6,7,8 ≥ 0
Thus, we have standard form:
x4 − 5x5 + 5x6 − 7x7 + 7x8 − 2 → min
Such that


5x4 − 2x5 + 2x6 + 6x7 − 6x8 ≥ 15






 3x4 + 4x5 − 4x6 − 9x7 + 9x8 ≥ 9
D : −3x4 − 4x5 + 4x6 + 9x7 − 9x8 ≥ −9




−7x4 − 3x5 + 3x6 − 5x7 + 5x8 ≥ −23




x
≥0
4,5,6,7,8
Exercise 7
Let x1 = x3 − x4 with x3,4 ≥ 0
Then we have canonnical form:
3x2 − 6x3 + 6x4 → min
Such that
D:



 5x2 + 2x3 + 2x4 = 10
−2x2 − 3x3 + 3x4 = 40



x2 ≥ 4, x3,4 ≥ 0
Exercise 8
Let x1 = x4 − x5 and x2 = x6 − x7 where x4,5,6,7 ≥ 0
Then we have canonnical form:
−x3 + x4 − x5 − 12x6 + 12x7 → min
Such that
D:



−2x3 + 5x4 − 5x5 − x6 + x7 = 10
20x3 − 2x4 + 2x5 − x6 + x7 = 30



x3,4,5,6,7 ≥ 0
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