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Lösung 1.3:6

Aus Online Mathematik Brückenkurs 2

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If we call the radius of the metal can
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If we call the radius of the metal can ''r'' and its height ''h'', then we can determine the can's volume and area by using the figures below,
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<math>r</math>
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and its height
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<math>h</math>, then we can determine the can's volume and area by using the figures below.
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Volume = (area of the base). (height)
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<math>=\pi r^{2}h</math>
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Area= (area of the base)+(area of the cylindrical surface)
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<math>=\pi r^{2}+2\pi h</math>
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{{Displayed math||<math>\begin{align}
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\text{Volume} &= \text{(area of the base)}\cdot\text{(height)}\\[5pt]
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&= \pi r^2\cdot h\,,\\[10pt]
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\text{Area} &= \text{(area of the base)} + \text{(area of the cylindrical surface)}\\[5pt]
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&= \pi r^2 + 2\pi rh\,\textrm{.}
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\end{align}</math>}}
[[Image:1_3_6_1.gif|center]]
[[Image:1_3_6_1.gif|center]]
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The problem can then be formulated as: minimise the can's area, <math>A = \pi r^2 + 2\pi h</math>, whilst at the same time keeping the volume, <math>V = \pi r^2h\,</math>, constant.
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The problem can then be formulated as: minimise the can's area,
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From the formula for the volume, we can make ''h'' the subject,
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<math>A=\pi r^{2}+2\pi h</math>, whilst at the same time keeping the volume,
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<math>V=\pi r^{2}h</math>, constant.
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From the formula for the volume, we can make
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{{Displayed math||<math>h=\frac{V}{\pi r^2}</math>}}
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<math>h</math>
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the subject,
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and express the area solely in terms of the radius, ''r'',
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<math>h=\frac{V}{\pi r^{2}}</math>
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{{Displayed math||<math>A = \pi r^2 + 2\pi r\cdot\frac{V}{\pi r^2} = \pi r^2 + \frac{2V}{r}\,\textrm{.}</math>}}
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and express the area solely in terms of the radius,
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<math>r</math>:
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<math>A=\pi r^{2}+2\pi r\bullet \frac{V}{\pi r^{2}}=\pi r^{2}+\frac{2V}{r}</math>
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The minimisation problem is then:
The minimisation problem is then:
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to minimise the area
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::Minimise the area <math>A(r) = \pi r^2 + \frac{2V}{r}</math>, when <math>r>0\,</math>.
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<math>A\left( r \right)=\pi r^{2}+\frac{2V}{r}</math>
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A(r)=..., when
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<math>r>0</math>.
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The area function
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The area function <math>A(r)</math> is differentiable for all <math>r>0</math> and the region of definition <math>r>0</math> has no endpoints (<math>r=0</math> does not satisfy <math>r>0</math>), so the function can only assume extreme values at critical points.
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<math>A\left( r \right)</math>
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is differentiable for all
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<math>r>0</math>
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and the region of definition
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<math>r>0</math>
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has no endpoints
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(
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<math>r=0\text{ }</math>
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does not satisfy
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<math>r>0</math>
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), so the function can only assume extreme values at critical points.
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The derivative is given by
The derivative is given by
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{{Displayed math||<math>A'(r) = 2\pi r - \frac{2V}{r^2}\,,</math>}}
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<math>{A}'\left( r \right)=2\pi r-\frac{2V}{r^{2}}</math>,
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and if we set the derivative equal to zero, so as to obtain the critical points, we get
and if we set the derivative equal to zero, so as to obtain the critical points, we get
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& 2\pi r-\frac{2V}{r^{2}}=0\quad \Leftrightarrow \quad 2\pi r=\frac{2V}{r^{2}} \\
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& 2\pi r - \frac{2V}{r^2} = 0\quad \Leftrightarrow \quad 2\pi r = \frac{2V}{r^2}\\[5pt]
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& \Leftrightarrow \quad r^{3}=\frac{V}{\pi }\quad \Leftrightarrow \quad r=\sqrt[3]{\frac{V}{\pi }} \\
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&\quad\Leftrightarrow \quad r^3=\frac{V}{\pi}\quad \Leftrightarrow \quad r=\sqrt[\scriptstyle 3]{\frac{V}{\pi}}\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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For this value of ''r'', the second derivative,
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{{Displayed math||<math>A''(r) = 2\pi + \frac{4V}{r^3}\,,</math>}}
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For this value of
 
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<math>r</math>, the second derivative,
 
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<math>{A}''\left( r \right)=2\pi +\frac{4V}{r^{3}}</math>,
 
has the value
has the value
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{{Displayed math||<math>A''\bigl(\sqrt[3]{V/\pi}\bigr) = 2\pi + \frac{4V}{V/\pi } = 6\pi > 0\,,</math>}}
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<math>{A}''\left( \sqrt[3]{\frac{V}{\pi }} \right)=2\pi +\frac{4V}{\frac{V}{\pi }}=6\pi >0</math>,
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which shows that <math>r=\sqrt[3]{V/\pi}</math> is a local minimum.
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which shows that
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Because the region of definition, <math>r>0</math>, is open (the endpoint
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<math>r=\sqrt[3]{{V}/{\pi }\;}</math>
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<math>r=0\text{ }</math> is not included) and unlimited, we cannot directly say that the area is least when <math>r = \sqrt[3]{V/\pi}\,</math>; it could be the case that area becomes smaller when <math>r\to 0</math> or <math>r\to \infty </math>. In this case, however, the area increases without bound as
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is a local minimum.
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<math>r\to 0</math> or <math>r\to \infty </math>, so <math>r=\sqrt[3]{V/\pi}</math>
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Because the region of definition,
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<math>r>0</math>, is open (the endpoint
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<math>r=0\text{ }</math>
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is not included) and unlimited, we cannot directly say that the area is least when
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<math>r=\sqrt[3]{{V}/{\pi }\;}</math>; it could be the case that area becomes smaller when
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<math>r\to 0</math>
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or
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<math>r\to \infty </math>. In this case, however, the area increases without bound as
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<math>r\to 0</math>
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or
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<math>r\to \infty </math>, so
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<math>r=\sqrt[3]{{V}/{\pi }\;}</math>
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really is a global minimum.
really is a global minimum.
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The metal can has the least area for a given volume
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The metal can has the least area for a given volume <math>V</math> when
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<math>V</math>
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when
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<math>r=\sqrt[3]{{V}/{\pi }\;}</math>, and
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<math>h=\frac{V}{\pi r^{2}}=\frac{V}{\pi }\left( \frac{V}{\pi } \right)^{-{2}/{3}\;}=\left( \frac{V}{\pi } \right)^{1-{2}/{3}\;}=\left( \frac{V}{\pi } \right)^{{1}/{3}\;}=\sqrt[3]{\frac{V}{\pi }}</math>
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{{Displayed math||<math>\begin{align}
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r &= \sqrt[3]{V/\pi}\,,\quad\text{and}\\[5pt]
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h &= \frac{V}{\pi r^{2}} = \frac{V}{\pi}\Bigl(\frac{V}{\pi}\Bigr)^{-2/3} = \Bigl( \frac{V}{\pi}\Bigr)^{1-2/3} = \Bigl(\frac{V}{\pi}\Bigr)^{1/3} = \sqrt[3]{\frac{V}{\pi}}\,\textrm{.}
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\end{align}</math>}}

Version vom 10:55, 21. Okt. 2008

If we call the radius of the metal can r and its height h, then we can determine the can's volume and area by using the figures below,

VolumeArea=(area of the base)(height)=r2h=(area of the base)+(area of the cylindrical surface)=r2+2rh.

The problem can then be formulated as: minimise the can's area, A=r2+2h, whilst at the same time keeping the volume, V=r2h, constant.

From the formula for the volume, we can make h the subject,

h=Vr2

and express the area solely in terms of the radius, r,

A=r2+2rVr2=r2+r2V.

The minimisation problem is then:

Minimise the area A(r)=r2+r2V, when r0.

The area function A(r) is differentiable for all r0 and the region of definition r0 has no endpoints (r=0 does not satisfy r0), so the function can only assume extreme values at critical points.

The derivative is given by

A(r)=2r−r22V

and if we set the derivative equal to zero, so as to obtain the critical points, we get

2r−r22V=02r=r22Vr3=Vr=3V.

For this value of r, the second derivative,

A(r)=2+r34V

has the value

A3V=2+4VV=60 

which shows that r=3V  is a local minimum.

Because the region of definition, r0, is open (the endpoint r=0 is not included) and unlimited, we cannot directly say that the area is least when r=3V ; it could be the case that area becomes smaller when r0 or r. In this case, however, the area increases without bound as r0 or r, so r=3V  really is a global minimum.

The metal can has the least area for a given volume V when

rh=3Vand=Vr2=VV−23=V1−23=V13=3V. 
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:6“