This is due tomorrow and i have answers but i think ive done it wrong! please help
1 answer:
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The problem can be solved step by step, if we know certain basic rules of summation. Following rules assume summation limits are identical.
Armed with the above rules, we can split up the summation into simple terms:
=> (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
Armed with the above rules, we can split up the summation into simple terms:
=> (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