An equilateral triangle and a square have the same perimeter. One side of the triangle measures 12 cm.
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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