linear programming. consider the following problem: a tailor has the following materials available...
TRANSCRIPT
Consider the following problem: A tailor has the following materials available in stock: 16 m2 of cotton, 11 m2 of silk and 15 m2 of wool. A suit requires 2 m2 of cotton, 1 m2 of silk and 1 m2 of wool to produce, while a gown requires 1 m2 of cotton, 2 m2 of silk and 3 m2 of wool. A clothing outlet will pay the tailor $150 for each suit and $250 for each gown he produces. How many suits and how many gowns should the tailor prepare in order to maximize his revenue from the sale of these articles of clothing?
y > mx + b lies above the line
x
y
1
b = –4 m = 3
test point (0, 0)
LS = y = 0
RS = 3(0) – 4 RS = –4 LS > RS
(0, 0) is in the region
y ≥ 3x – 4
y < mx + b lies below the line
The line is in the region as well
RS = 3(x) – 4
–4
3
If we solve the inequality for y we get y > mx + b (above the line)
x
y
–2
a = 5 b = –2
test point (0, 0)
LS = 0
RS = 10 LS < RS
(0, 0) is in the region
2x – 5y < 10
5
The line is not in the region
x > a lies to the right of the line
x
y
2
a = 2
vertical line
test point (0, 0) LS = 0
RS = 6
LS < RS
(0, 0) is not in the region
3x > 6
x < a lies to the left of the line
x
y
–2
x – y 2
2x + y –2
y 1
2
1
–1
Above this line
Below this line
Above this lineThe boundary lines are included
y
x
PointIntercept Position
SlopeInclinationDirection
Consider the straight line on the x-y plane
y = mx + b
y = m(x a)
Ax + By + C = 0
m is called a parameter (or free variable)
y = 3x + b y = a(x – h)2 + k
x2 + y2 = r2
Here are some additional examples:
y = asin(x – p) + q
y = mx + 2
y = logbx + c
y
x
The parameter k is part of an expression that determines the y-intercept (or the x-intercept).
3x – 4y = k
The slope is always ¾
Definitions:The objective function:
The feasible region:
The corner points:
A linear function (consisting of two variables if a plane is used) which is to be maximized or minimized.
The set of interior and boundary points of a polygon defined by the inequality relations which specify the restrictions (the constraints) on the variables used.
The vertices of the feasible region (the set of possible solutions to the problem).
The theory of linear programming states that, to find the optimum value of f, where f = Ax + By, we need only test the values of f at the corner points of the feasible region R. This assumes that region R is a convex set of points determined by the intersection of 3 or more linear inequalities. The region R may be open (also called bounded) or closed (also called unbounded). The region is usually open when we are testing for a minimum value of f.
Maximize: 150x + 250yR =
Products Resources
Resources Suits (x) Gowns (y) Available
Cotton 2 1 16
Silk 1 2 11
Wool 1 3 15
Product - Resource Chart (figures in m2)
2x + y = 16 [1]
x + 3y = 15 [2]
6x + 3y = 48 [3]3 [1]
[3] [2] 5x = 33
x = 6.6
y = 2.8 Substituting in [1]
Divide by 5
Thus, (x, y) = (6.6, 2.8)
Point E in previous diagram
Conclusion: The maximum revenue can be generated by producing 7 suits and 2 gowns
Table for Testing the Revenue Function
Corner Point Objective Function: 150x + 250y
A ( 0 , 5 ) R = 150(0) + 250(5) B ( 3 , 4 ) R = 150(3) + 250(4) = 1450 C ( 7 , 2 ) R = 150(7) + 250(2) = 1550 D ( 8 , 0 ) R = 150(8) + 250(0) = 1200 O (0, 0) R = 150(0) + 250(0) = 0
= 1250
= 1550