Answer:
The sum of all possible values of n is 9.
Step-by-step explanation:
We are going to solve this problem by subtracting areas.
For the first stage, the rectangular area of the formation is :
[tex](n-2).(n+8)[/tex]
In the second stage, the rectangular area of the formation is :
[tex]n(2n-3)[/tex]
We know that in this second formation they excluded all the drummers and also we know that there are at least 4 drummers.
Therefore, the difference between the areas of the first and the second formation is :
[tex](n-2).(n+8)-n.(2n-3)[/tex] and this area must be at least 4 (because of the drummers excluded)
[tex](n-2).(n+8)-n.(2n-3)\geq 4[/tex]
[tex]n^{2}+8n-2n-16-2n^{2}+3n\geq 4[/tex]
[tex]-n^{2}+9n-16\geq 4[/tex]
[tex]-n^{2}+9n-20\geq 0[/tex] (I)
We need to solve this and find the possibles ''n'' that satisfy the inequality.
First we look for the values that satisfy
[tex]-n^{2}+9n-20=0[/tex] (II)
Using the quadratic equation :
[tex]n_{1}=4\\n_{2}=5[/tex]
For this values of ''n'' the inequality (I) is satisfied.
Now we study the vertex.
Given a quadratic function [tex]f(x)=ax^{2}+bx+c[/tex]
The coordinate ''x'' of the vertex is [tex]\frac{-b}{2a}[/tex]
For (II)
[tex]a=-1\\b=9\\c=-20[/tex]
[tex]\frac{-b}{2a}=\frac{-9}{2(-1)}=\frac{9}{2}=4.5[/tex]
This is the coordinate ''x'' of the vertex.
For the coordinate ''y'' we calculate [tex]f(xVertex)[/tex]
[tex]f(4.5)=-(4.5)^{2}+9(4.5)-20=0.25[/tex]
That is positive. The coordinates of the vertex are [tex](4.5,0.25)[/tex]
In the quadratic function [tex]a=-1\\a<0[/tex]
So it is a negative quadratic function.
We conclude that for the interval
[4,5] the quadratic function is positive, therefore between [4,5] the inequality (I) is satisfied.
The two possible values for n are 4 and 5.
Finally, [tex]4+5=9[/tex] is the sum of all possible values of n
(Notice that n must be an integer number)
