A ladder of length 12.5m leans against a wall of height 6.4m. Let x m be the horizontal distance that the ladder ectends beyond the wall. Find the maximum value of x.
Absolute Extrema & Optimization Problems
2007-05-29 9:05 am
回答 (2)
2007-05-29 7:15 pm
✔ 最佳答案
As follows:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May07/Crazydiff62.jpg
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/May07/Crazydiff63.jpg
參考: My Maths knowledge
2007-05-29 6:57 pm
Nice illustration!
Let t=angle between ladder and vertical,
then
summing the length of the ladder formed by the hypothenuses of the two triangles,
x/sin(t)+6.4/cos(t)=12.5
solving,
x=12.5sin(t)-6.4tan(t)
dx/dt=12.5cos(t)-6.4sec^2(t)
For x to have maximum or minimum,
dx/dt=0 =>
dx/dt=12.5cos(t)-6.4sec^2(t)=0
from which the real root of
t=arccos(4/5) can be found.
Substituting, the sides of the lower triangle=6.4, 4.8 and 8
The sides of the upper triangle=(12.5-8)=4.5, 3.6 and 2.7
Thus the maximum value of x=2.7.
Final step, show the value is a maximum:
d2x/dt2=-12.8*sec(t)^2*tan(t)-12.5*sin(t)
for t=acos(4/5),
d2x/dt2=-12.8*sec(t)^2*tan(t)-12.5*sin(t)=-22.5 <0, therefore a maximum.
Let t=angle between ladder and vertical,
then
summing the length of the ladder formed by the hypothenuses of the two triangles,
x/sin(t)+6.4/cos(t)=12.5
solving,
x=12.5sin(t)-6.4tan(t)
dx/dt=12.5cos(t)-6.4sec^2(t)
For x to have maximum or minimum,
dx/dt=0 =>
dx/dt=12.5cos(t)-6.4sec^2(t)=0
from which the real root of
t=arccos(4/5) can be found.
Substituting, the sides of the lower triangle=6.4, 4.8 and 8
The sides of the upper triangle=(12.5-8)=4.5, 3.6 and 2.7
Thus the maximum value of x=2.7.
Final step, show the value is a maximum:
d2x/dt2=-12.8*sec(t)^2*tan(t)-12.5*sin(t)
for t=acos(4/5),
d2x/dt2=-12.8*sec(t)^2*tan(t)-12.5*sin(t)=-22.5 <0, therefore a maximum.
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