The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
[tex]\sum{a+b}=\sum{a}+\sum{b}[/tex]
[tex]\sum{kx}=k\sum{x}[/tex]
[tex]\sum_{r=1}^n{1}=n[/tex]
[tex]\sum_{r=1}^n{r}=n(n+1)/2[/tex]
Armed with the above rules, we can split up the summation into simple terms:
[tex]\sum_{r=1}^n{40r-21n+8}=n[/tex]
[tex]=40\sum_{r=1}^n{r}-21n\sum_{r=1}^n{1}+8\sum_{r=1}^n{1}[/tex]
[tex]=40\frac{n(n+1)}{2}-21n^2+8n[/tex]
[tex]=20n(n+1)-21n^2+8n[/tex]
[tex]=28n-n^2[/tex]
=> (a) f(x)=28n-n^2
=> f'(x)=28-2n
=> at f'(x)=0 => x=14
Since f''(x)=-2 <0 therefore f(14) is a maximum
(b) f(x) is a maximum when n=14
(c) the maximum value of f(x) is f(14)=196
