Lösung 1.3:5

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The area of the cross-section is
The area of the cross-section is
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{{Displayed math||<math>\begin{align}
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A(\alpha) &= 10\cdot 10\cos\alpha + 2\cdot\frac{1}{2}\cdot 10\cos\alpha \cdot 10\sin\alpha\\[5pt]
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&= 100\cos \alpha (1+\sin\alpha)\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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If we limit the angle to lie between <math>0</math> and <math>\pi/2</math>, the problem can be formulated as:
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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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::Maximise the function <math>A(\alpha) = 100\cos\alpha (1+\sin\alpha)</math> when
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<math>0\le \alpha \le {\pi }/{2}\,</math>.
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If we limit the angle to lie between
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The area function is a differentiable function and the area is least when <math>\alpha=0</math> or <math>\alpha=\pi/2\,</math>, so the area must assume its maximum at a critical point of the area function.
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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:
We differentiate the area function:
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{{Displayed math||<math>\begin{align}
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A'(\alpha) &= 100\cdot (-\sin\alpha)\cdot (1+\sin\alpha) + 100\cdot\cos\alpha \cdot \cos\alpha\\[5pt]
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&= -100\sin\alpha - 100\sin^2\!\alpha + 100\cos^2\!\alpha\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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At a critical point <math>A'(\alpha)=0</math> and this gives us the equation
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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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{{Displayed math||<math>\sin\alpha + \sin^2\!\alpha - \cos^2\!\alpha = 0</math>}}
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<math>\begin{align}
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after eliminating the factor -100. We replace <math>\cos^2\!\alpha</math> with <math>1-\sin^2\!\alpha</math> (according to the Pythagorean identity) to obtain an equation solely in <math>\sin\alpha\,</math>,
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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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{{Displayed math||<math>\begin{align}
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<math>\sin \alpha =-\frac{1}{4}\pm \frac{3}{4}</math>
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\sin\alpha + \sin^2\!\alpha - (1-\sin^2\!\alpha) &= 0\\[5pt]
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i.e.
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2\sin^2\!\alpha + \sin\alpha - 1 &= 0\,\textrm{.}
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<math>\sin \alpha =-1</math>
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\end{align}</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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This is a second-degree equation in <math>\sin\alpha</math> and completing the square gives that
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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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{{Displayed math||<math>\begin{align}
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2\bigl(\sin\alpha + \tfrac{1}{4}\bigr)^{2} - 2\bigl(\tfrac{1}{4}\bigr)^2 - 1 &= 0\\[5pt]
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\bigl(\sin\alpha + \tfrac{1}{4}\bigr)^2 &= \frac{9}{16}
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\end{align}</math>}}
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If we summarize, we know therefore that the cross-sectional area has local minimum points at
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and we obtain <math>\sin\alpha = -\tfrac{1}{4}\pm \tfrac{3}{4}</math>, i.e. <math>\sin \alpha = -1</math> or <math>\sin \alpha = \tfrac{1}{2}\,</math>.
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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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The case <math>\sin \alpha =-1</math> is not satisfied for <math>0\le \alpha \le \pi/2</math> and <math>\sin \alpha = \tfrac{1}{2}</math> gives <math>\alpha = \pi/6</math>. Thus <math>\alpha = \pi/6\,</math> is a critical point.
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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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If we summarize, we know therefore that the cross-sectional area has local minimum points at <math>\alpha = 0</math> and <math>\alpha = \pi/2</math> and that we have a critical point at <math>\alpha = \pi/6\,</math>. This critical point must be a maximum, which we can also show using the second derivative,
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which is negative at
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{{Displayed math||<math>\begin{align}
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<math>\alpha ={\pi }/{6}\;</math>,
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A''(\alpha) &= -100\cos\alpha - 100\cdot 2\sin\alpha\cdot\cos\alpha + 100\cdot 2\cos\alpha \cdot (-\sin\alpha)\\[5pt]
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&= -100\cos\alpha (1+4\sin\alpha)\,,
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\end{align}</math>}}
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which is negative at <math>\alpha = \pi/6</math>,
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<math>\begin{align}
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{{Displayed math||<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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A''(\pi/6) &= -100\cos\frac{\pi}{6}\cdot \Bigl(1+4\sin\frac{\pi}{6}\Bigr)\\[5pt]
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& =-100\centerdot \frac{\sqrt{3}}{2}\centerdot \left( 1+4\centerdot \frac{1}{2} \right)<0 \\
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&= -100\cdot\frac{\sqrt{3}}{2}\cdot \Bigl( 1+4\cdot \frac{1}{2} \Bigr)<0\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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There are no local maximum points other than
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There are no local maximum points other than <math>\alpha = \pi/6\,</math>, which must therefore also be a global maximum.
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<math>\alpha ={\pi }/{6}\;</math>, which must therefore also be a global maximum.
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Version vom 10:30, 21. 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

\displaystyle \begin{align}

A(\alpha) &= 10\cdot 10\cos\alpha + 2\cdot\frac{1}{2}\cdot 10\cos\alpha \cdot 10\sin\alpha\\[5pt] &= 100\cos \alpha (1+\sin\alpha)\,\textrm{.} \end{align}

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

Maximise the function \displaystyle A(\alpha) = 100\cos\alpha (1+\sin\alpha) when

\displaystyle 0\le \alpha \le {\pi }/{2}\,.

The area function is a differentiable function and the area is least when \displaystyle \alpha=0 or \displaystyle \alpha=\pi/2\,, so the area must assume its maximum at a critical point of the area function.

We differentiate the area function:

\displaystyle \begin{align}

A'(\alpha) &= 100\cdot (-\sin\alpha)\cdot (1+\sin\alpha) + 100\cdot\cos\alpha \cdot \cos\alpha\\[5pt] &= -100\sin\alpha - 100\sin^2\!\alpha + 100\cos^2\!\alpha\,\textrm{.} \end{align}

At a critical point \displaystyle A'(\alpha)=0 and this gives us the equation

\displaystyle \sin\alpha + \sin^2\!\alpha - \cos^2\!\alpha = 0

after eliminating the factor -100. We replace \displaystyle \cos^2\!\alpha with \displaystyle 1-\sin^2\!\alpha (according to the Pythagorean identity) to obtain an equation solely in \displaystyle \sin\alpha\,,

\displaystyle \begin{align}

\sin\alpha + \sin^2\!\alpha - (1-\sin^2\!\alpha) &= 0\\[5pt] 2\sin^2\!\alpha + \sin\alpha - 1 &= 0\,\textrm{.} \end{align}

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

\displaystyle \begin{align}

2\bigl(\sin\alpha + \tfrac{1}{4}\bigr)^{2} - 2\bigl(\tfrac{1}{4}\bigr)^2 - 1 &= 0\\[5pt] \bigl(\sin\alpha + \tfrac{1}{4}\bigr)^2 &= \frac{9}{16} \end{align}

and we obtain \displaystyle \sin\alpha = -\tfrac{1}{4}\pm \tfrac{3}{4}, i.e. \displaystyle \sin \alpha = -1 or \displaystyle \sin \alpha = \tfrac{1}{2}\,.

The case \displaystyle \sin \alpha =-1 is not satisfied for \displaystyle 0\le \alpha \le \pi/2 and \displaystyle \sin \alpha = \tfrac{1}{2} gives \displaystyle \alpha = \pi/6. Thus \displaystyle \alpha = \pi/6\, is a critical point.


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

\displaystyle \begin{align}

A''(\alpha) &= -100\cos\alpha - 100\cdot 2\sin\alpha\cdot\cos\alpha + 100\cdot 2\cos\alpha \cdot (-\sin\alpha)\\[5pt] &= -100\cos\alpha (1+4\sin\alpha)\,, \end{align}

which is negative at \displaystyle \alpha = \pi/6,

\displaystyle \begin{align}

A''(\pi/6) &= -100\cos\frac{\pi}{6}\cdot \Bigl(1+4\sin\frac{\pi}{6}\Bigr)\\[5pt] &= -100\cdot\frac{\sqrt{3}}{2}\cdot \Bigl( 1+4\cdot \frac{1}{2} \Bigr)<0\,\textrm{.} \end{align}

There are no local maximum points other than \displaystyle \alpha = \pi/6\,, which must therefore also be a global maximum.

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