SOLUTION: Suppose a rope is to be strung from the tops of two posts of heights 10 feet and 8 feet that are 24 feet apart. The rope needs to be staked to the ground between the two posts. W

Question 34647: Suppose a rope is to be strung from the tops of two posts of heights 10 feet and 8 feet that are 24 feet apart. The rope needs to be staked to the ground between the two posts. What is the shortest rope that can possibly be used? How far from the taller post should it be staked?
Answer by venugopalramana(3286) (Show Source): You can 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
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