+------------+---
| y / |
| / |b ft
| / |
| /+------
| / |
|x / |
| / |
| / |
| / |
| / |
| / |
|/ |
+<-a ft-->|
In the above diagram (x-b)/a = x/y (by similar triangles).
Therefore y = ax/(x-b).The length of the diagonal line L is given by: L² = x²+y² = x²+a²*x²/(x-b)².
The minimum value of L is the length of the largest panel. L is minimal when L² is minimal, i.e., 2L.dL/dx = 2x+2a².x/(x-b)²-2a²x²/(x-b)³ = 0. This can be simplified into (x-b)³+a²*(x-b) = a²x.
Then x = a^(2/3).b^(1/3)+b = b^(1/3).(a^(2/3)+b^(2/3)).
Then y = a.(a^(2/3).b^(1/3)+b)/a^(2/3)*b^(1/3) = a^(1/3).b^(2/3)+a =
a^(1/3).(a^(2/3)+b^(2/3)).
This yields the result
L = √(a^(2/3)+b^(2/3))*(a^(2/3)+b^(2/3)) = √(a^(2/3)+b^(2/3))³.
This means that the longest panel has length √(a^(2/3)+b^(2/3))³
NOTES: The sign of the second derivative must be studied to be sure that we do have a minimum. Also, as x and y are bounded (x>b, y>a) we could have an extreme at x=b or y=a; they should be studied separately. |
