<span>I think you know by now that I strongly encourage everyone to shoot a proper round and whatever the score is, to submit it to our Records Officer, Giles Conn. Think of it as an annual competition (a) to wear him out, and (b) to see if we can altogether, beat last year's tally. Also, for the outdoor season rounds, you can have a go at achieving the St Wilfred trophy. I've won it 3 years running (last year jointly with Terry Skinner), but they wouldn't let me keep it this time, sadly.</span>
Answer:
Prove what is similar?????
If A=w(50-w)
A=50w-w^2
dA/dw=50-2w
d2A/dw2=-2
Since the acceleration is a constant negative, that means that when velocity, dA/dw=0, it is at an absolute maximum for A(w)...
dA/d2=0 only when 50=2w, w=25
So as the case with any rectangle, the perfect square will enclose the greatest area possible with respect to a given amount of material to enclose that area...
So the greatest area occurs when W=L=25 in this case:
A(25)=50w-w^2
Area maximum is thus:
Amax=50(25)-(25)^2=625 u^2