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put this solution on YOUR website!LET THE 2 POLES BE AB AND CD WITH A AND C ON GROUND
AB=10.....CD=8.......AC=24....LET THE STAKE POINT BE P ON THE
GROUND....LET AP=X...SO....PC=24-X
LENGTH OF ROPE=L=BP+PD.
L=SQRT{AB^2+BP^2}+SQRT{CD^2+PC^2}.....
L=SQRT{(100+X^2)+SQRT{64+(24-X)^2}
DL/DX=0.5*2X/(100+X^2)^(0.5)+0.5*2*(-1)(24-X)/{64+(24-X)^2}^(0.5)=0.
FOR MINIMUM L...... DL/DX=0....SO
X/(100+X^2)^0.5=(24-X)/{64+(24-X)^2}^0.5....SQUARING AND CROSS
MULTIPLYING WE GET
X^2{64+(24-X)^2}=(24-X)^2(100+X^2)
64X^2=100(24-X)^2....TAKING SQUARE ROOT
10(24-X)=+8X.....OR........-8X
240-10X=8X
18X=240
X=40/3..........................I
IF WE TAKE -8X...THEN
240-10X=-8X
2X=240
X=120...THIS IS NOT POSSIBLE INCE AC=24 ONLY AND X CANT BE MORE THAN THAT
SO
X=40/3...IS THE STAKE POINT FROM LONGER POLE...AB
MINIMUM LENGTH OF ROPE =PUT X=40/3 IN
L=SQRT{(100+X^2)+SQRT{64+(24-X)^2}=30
