Question 1203027: The effectiveness of a television commercial depends on how many times a viewer watches it.After some experiments, an advertising agency found that if the effectiveness E is measured on a scale of 0 to 10,then E(n)=2/3×n-1/90×n^2, where n is the number of times a viewer watches a given commercial .By completing the square of E(n),find how many times a viewer should watch the commercial for a commercial to have maximum effectiveness?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849)   (Show Source):
Answer by ikleyn(52756)  (Show Source):
You can put this solution on YOUR website! .
The effectiveness of a television commercial depends on how many times a viewer watches it.
After some experiments, an advertising agency found that if the effectiveness E
is measured on a scale of 0 to 10, then E(n)=2/3×n-1/90×n^2, where n is the number
of times a viewer watches a given commercial.
By completing the square of E(n), find how many times a viewer should watch
the commercial for a commercial to have maximum effectiveness?
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For this problem, @MathLover1 in her post responded first using Calculus,
which is irrelevant to the request.
Then she announced that she is going to complete the square,
but that part of her post is incorrect or unreadable, at all.
By knowing empirically that it is impossible to get an adequate answer from this woman,
I came to make the job in a way how it SHOULD be done.
E(n) =  write everything with the common denominator 
=  take the factor  out of parentheses
=  transform equivalently
=  +  =  + 10.
Thus completing the square is done.
The term is the parabola opened downward.
The addend 10 shifts the parabola 10 units vertically up.
The maximum of 10 is achieved at n = 30. ANSWER
Solved.
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On completing the square, see this lesson
- HOW TO complete the square - Learning by examples
in this site.
Learn the subject from there once and for all.
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