:)
In 6.3, we learned that what kind of numbers were dealing with contribute to our answer. This still applies to 6.4.
In 6.4, we learned about optimization problems. In an optimization problem, it asks you for the maximum and minimum possible solution.
Optimization problem
a problem where a quantity must be maximized or minimized following a set of guidlines or conditions.
Constraint
A limiting condition of the optimization problem being modelled, represented by a linear inequality.
Objective function
in an optimization problem, the equation that represents the relationship between the two variables in the system of linear inequalities and the quantity to be optimized.
Feasible region
The solution for a system of linear inequalities that is modelling an optimization problem.
The area in yellow represents the feasible region.
When graphins linear inequalities we must remember which part of the inequality goes on certain areas of the graph.
We then went over the last page of our 6.4 booklet.
We found out that you can tell you're dealing with an optimization problem if you have to find the minimum or maximum.
We found that the constraints are the linear inequalities in the problem.
We did practice on pages 330-331 # 2,3,5,6.
In 6.5 we looked at the race car and suv problem again.
We realised that the maximum and minimum was on the vertices of the feasible region, but we had to find out which two vertices.
The objective function to optiimize was: C = 8r + 12s
C = cost
r = race cars
s = suvs
We found that (60, 40) was the maximum.
C = 12(60) + 8(40)
C= $1040
We found that (30, 40) was the minimum.
C = 12(30) + 8(40)
C = $680
But does the minimum cost satisfy all the constraints?
Constraints:
r<40 (less than or equal to)
s<60 (less than or equal to)
s+r>70 (greater than or equal to)
YES IT SATISFIES ALL CONSTRAINTS!
We then did practice questions on pages 334-335 #1-3
here is a math video to brighten your day! lolol
http://www.youtube.com/watch?v=cgEuUzHYvOY">
Next will be Aaron :D
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