Lösung 1.3:7
Aus Online Mathematik Brückenkurs 2
The whole procedure can be illustrated by the figure below:
Because it is the cornet's volume we want to maximise, it is appropriate to start by introducing some notation for the dimensions of the cornet.
With these dimensions, the volume of the cornet will be the same as that of a cone,
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To go further, we now need to express the radius
When we cut out a circular sector of angle −
)R
On the other hand, the cornet's upper circular edge has a circumference
r
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We have thus managed to express the radius
In order to obtain the height
This means that
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Hence, we have expressed
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At last, we can mathematically formulate the problem:
- Maximise
V( , where)=31
R3
2
2
−
2
1−
2
2
−
2
0 .2
- Maximise
Before we start to try and solve this problem, we can observe that the variable
−
)
2
−
)
2
- Maximise
V(x)=31 , whenR3x2
1−x2
0 .x
1
- Maximise
When either
We differentiate,
\displaystyle V'(x) = \frac{1}{3}\pi R^3\cdot 2x\cdot \sqrt{1-x^2} + \frac{1}{3}\pi R^3x^2\cdot\frac{1}{2\sqrt{1-x^2}}\cdot (-2x)\,, |
and begin simplifying this expression. The strategy is to try to take out as many factors as possible, so that we see more easily when some factor, and hence the derivative, becomes zero,
\displaystyle \begin{align}
V'(x) &= \frac{2}{3}\pi R^3x\sqrt{1-x^2} - \frac{1}{3}\pi R^3x^3\frac{1}{\sqrt{1-x^2}}\\[5pt] &= \frac{1}{3}\pi R^3\frac{x}{\sqrt{1-x^2}}\bigl[ 2(1-x^2)-x^2\bigr]\\[5pt] &= \frac{1}{3}\pi R^3\frac{x}{\sqrt{1-x^2}}(2-3x^2)\,\textrm{.} \end{align} |
The derivative is zero when \displaystyle x=0 (which is an endpoint) or when \displaystyle 2-3x^2=0, i.e. \displaystyle x=\sqrt{2/3}\,. (The point \displaystyle x=-\sqrt{2/3} lies outside \displaystyle 0\le x\le 1.)
With the help of a table of the sign of the factors,
\displaystyle x | \displaystyle 0 | \displaystyle \sqrt{\tfrac{2}{3}} | \displaystyle 1 | ||
\displaystyle x | \displaystyle 0 | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + |
\displaystyle \sqrt{1-x^2} | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle + | \displaystyle 0 |
\displaystyle 2-3x^2 | \displaystyle + | \displaystyle + | \displaystyle 0 | \displaystyle - | \displaystyle - |
we can write down a table of the sign of the derivative itself,
\displaystyle x | \displaystyle 0 | \displaystyle \sqrt{\tfrac{2}{3}} | \displaystyle 1 | ||
\displaystyle V'(x) | \displaystyle 0 | \displaystyle + | \displaystyle 0 | \displaystyle - | |
\displaystyle V(x) | \displaystyle 0 | \displaystyle \nearrow | \displaystyle \tfrac{4}{9\sqrt{3}}\pi R^3 | \displaystyle \searrow | \displaystyle 0 |
and see that \displaystyle x=\sqrt{2/3} is a global maximum. The value \displaystyle x = \sqrt{2/3} corresponds to the \displaystyle \alpha-value
\displaystyle \sqrt{\frac{2}{3}}=\frac{2\pi-\alpha }{2\pi}\quad \Leftrightarrow\quad \alpha = 2\pi \bigl(1-\sqrt{2/3}\,\bigr)\ \text{radians.} |