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Lösung 1.3:5

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.3:5 moved to Solution 1.3:5: Robot: moved page)
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The channel holds most water when its cross-sectional area is greatest.
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[[Image:1_3_5_1_1.gif|center]]
[[Image:1_3_5_1_1.gif|center]]
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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.
[[Image:1_3_5_1_2.gif|center]]
[[Image:1_3_5_1_2.gif|center]]
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The area of the cross-section is
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<math>\begin{align}
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& A\left( \alpha \right)=10\centerdot 10\cos \alpha +2\centerdot \frac{1}{2}\centerdot 10\cos \alpha \centerdot 10\sin \alpha \\
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& =100\cos \alpha \left( 1+\sin \alpha \right) \\
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\end{align}</math>
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If we limit the angle to lie between
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<math>0</math>
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and
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<math>{\pi }/{2}\;</math>, the problem can be formulated as :
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Maximise the function
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<math>A\left( \alpha \right)=100\cos \alpha \left( 1+\sin \alpha \right)</math>
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when
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<math>0\le \alpha \le {\pi }/{2}\;</math>
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.
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The area function is a differentiable function and the area is least when
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<math>\alpha =0</math>
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or
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<math>\alpha ={\pi }/{2}\;</math>, so the area must assume its maximum at a critical point of the area function.
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We differentiate the area function:
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<math>\begin{align}
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& {A}'\left( \alpha \right)=100\centerdot \left( -\sin \alpha \right)\centerdot \left( 1+\sin \alpha \right)+100\centerdot \cos \alpha \centerdot \cos \alpha \\
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& =-100\sin \alpha -100\sin ^{2}\alpha +100\cos ^{2}\alpha \\
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\end{align}</math>
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At a critical point
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<math>{A}'\left( \alpha \right)=0</math>
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and this gives us the equation
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<math>\sin \alpha +\sin ^{2}\alpha -\cos ^{2}\alpha =0</math>
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after eliminating the factor
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<math>-100</math>. We replace
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<math>\cos ^{2}\alpha </math>
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with
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<math>1-\sin ^{2}\alpha </math>
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(according to the Pythagorean identity) to obtain an equation solely in
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<math>\sin \alpha </math>,
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<math>\begin{align}
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& \sin \alpha +\sin ^{2}\alpha -\left( 1-\sin ^{2}\alpha \right)=0 \\
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& 2\sin ^{2}\alpha +\sin \alpha -1=0 \\
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\end{align}</math>
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This is a second-degree equation in sin alpha and completing the square gives that
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<math>\begin{align}
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& 2\left( \sin \alpha +\frac{1}{4} \right)^{2}-2\left( \frac{1}{4} \right)^{2}-1=0 \\
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& \left( \sin \alpha +\frac{1}{4} \right)^{2}=\frac{9}{16} \\
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\end{align}</math>,
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and we obtain
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<math>\sin \alpha =-\frac{1}{4}\pm \frac{3}{4}</math>
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i.e.
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<math>\sin \alpha =-1</math>
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or
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<math>\sin \alpha =\frac{1}{2}</math>
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The case
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<math>\sin \alpha =-1</math>
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is not satisfied for
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<math>0\le \alpha \le {\pi }/{2}\;</math>
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and
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<math>\sin \alpha =\frac{1}{2}</math>
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gives
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<math>\alpha ={\pi }/{6}\;</math>. Thus
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<math>\alpha ={\pi }/{6}\;</math>
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is a critical point.
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If we summarize, we know therefore that the cross-sectional area has local minimum points at
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<math>\alpha =0</math>
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and
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<math>\alpha ={\pi }/{2}\;</math>
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and that we have a critical point at .
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<math>\alpha ={\pi }/{6}\;</math>
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This critical point must be a maximum, which we can also show using the second derivative,
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<math>\begin{align}
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& {A}''\left( \alpha \right)=-100\cos \alpha -100\centerdot 2\sin \alpha \centerdot \cos \alpha +100\centerdot 2\cos \alpha \centerdot \left( -\sin \alpha \right) \\
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& =-100\cos \alpha \left( 1+4\sin \alpha \right) \\
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\end{align}</math>
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which is negative at
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<math>\alpha ={\pi }/{6}\;</math>,
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<math>\begin{align}
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& {A}''\left( \frac{\pi }{6} \right)=-100\cos \frac{\pi }{6}\centerdot \left( 1+4\sin \frac{\pi }{6} \right) \\
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& =-100\centerdot \frac{\sqrt{3}}{2}\centerdot \left( 1+4\centerdot \frac{1}{2} \right)<0 \\
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\end{align}</math>
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There are no local maximum points other than
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<math>\alpha ={\pi }/{6}\;</math>, which must therefore also be a global maximum.

Version vom 10:31, 16. Okt. 2008

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


A=1010cos+22110cos10sin=100cos1+sin


If we limit the angle to lie between 0 and 2, the problem can be formulated as :

Maximise the function A=100cos1+sin  when 02 . The area function is a differentiable function and the area is least when =0 or =2, so the area must assume its maximum at a critical point of the area function.

We differentiate the area function:


A=100−sin1+sin+100coscos=−100sin−100sin2+100cos2 


At a critical point A=0  and this gives us the equation

sin+sin2−cos2=0


after eliminating the factor −100. We replace cos2 with 1−sin2 (according to the Pythagorean identity) to obtain an equation solely in sin,

sin+sin2−1−sin2=02sin2+sin−1=0 

This is a second-degree equation in sin alpha and completing the square gives that

2sin+412−2412−1=0sin+412=916,

and we obtain sin=−4143 i.e. sin=−1 or sin=21

The case sin=−1 is not satisfied for 02 and sin=21 gives =6. Thus =6

is a critical point.

If we summarize, we know therefore that the cross-sectional area has local minimum points at =0 and =2 and that we have a critical point at . =6 This critical point must be a maximum, which we can also show using the second derivative,


A=−100cos−1002sincos+1002cos−sin=−100cos1+4sin 


which is negative at =6,


A6=−100cos61+4sin6=−100231+4210 

There are no local maximum points other than =6, which must therefore also be a global maximum.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:5“