Lösung 1.3:5
Aus Online Mathematik Brückenkurs 2
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Version vom 10:12, 11. Mär. 2009
The channel holds most water when its cross-sectional area is greatest.
By dividing up the channel's cross-section into a rectangle and two triangles we can, with the help of a little trigonometry, determine the lengths of the rectangle's and triangles' sides, and thence the cross-sectional area.
The area of the cross-section is
 )=10 10cos+22110cos10sin=100cos(1+sin). |
If we limit the angle to lie between 
2
- Maximise the function
A( when)=100cos(1+sin)
- Maximise the function
2
The area function is a differentiable function and the area is least when =0=2
We differentiate the area function:
 ()=100(−sin)(1+sin)+100coscos=−100sin−100sin2+100cos2. |
At a critical point ()=0
| +sin2−cos2=0 |
after eliminating the factor -100. We replace
| +sin2−(1−sin2)2sin2+sin−1=0=0. |
This is a second-degree equation in
 sin+41 2−2412−1sin+412=0=916 |
and we obtain =−41
43=−1=21
The case =−12=21=6=6
If we summarize, we know therefore that the cross-sectional area has local minimum points at =0=2=6
()=−100cos−1002sincos+1002cos(−sin)=−100cos(1+4sin) |
which is negative at =6
(6)=−100cos6 1+4sin6 =−1002 31+421 0. |
There are no local maximum points other than =6
