Lösung 1.3:4

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If we call the
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If we call the ''x''-coordinate of the point <math>P</math> <math>x</math>, then its ''y''-coordinate is <math>1-x^{2}</math>, because <math>P</math> lies on the curve <math>y=1-x^{2}</math>.
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<math>x</math>
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-coordinate of the point
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<math>P</math>
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<math>x</math>, then its
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[[Image:1_3_4-1-1.gif|center]]
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<math>y</math>
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-coordinate is
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<math>1-x^{2}</math>, because
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<math>P</math>
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lies on the curve
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<math>y=1-x^{2}</math>.
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FIGURE
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The area of the rectangle is then given by
The area of the rectangle is then given by
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{{Displayed math||<math>A(x) = \text{(base)}\cdot\text{(height)} = x\cdot (1-x^2)</math>}}
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<math>A\left( x \right)=</math>
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and we will try to choose <math>x</math> so that this area function is maximised.
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base *height
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<math>=x\centerdot \left( 1-x^{2} \right)</math>.
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To begin with, we note that, because <math>P</math> should lie in the first quadrant, <math>x\ge 0</math> and also <math>y=1-x^2\ge 0</math>, i.e. <math>x\le 1</math>. We should therefore look for the maximum of <math>A(x)</math> when <math>0\le x\le 1\,</math>.
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and we will try to choose
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<math>x</math>
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so that this area function is maximised.
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To begin with, we note that, because
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<math>P</math>
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should lie in the first quadrant,
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<math>x\ge 0</math>
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and also
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<math>y=1-x^{2}\ge 0</math>, i.e.
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<math>x\le 1</math>. We should therefore look for the maximum of
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<math>A\left( x \right)</math>
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when
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<math>0\le x\le 1</math>.
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There are three types of points which can maximise the area function:
There are three types of points which can maximise the area function:
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1. critical points,
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# critical points,
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2. points where the function is not differentiable,
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# points where the function is not differentiable,
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3. endpoints of the region of definition.
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# endpoints of the region of definition.
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The function
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<math>A\left( x \right)=x\left( 1-x^{2} \right)</math>
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is differentiable everywhere, so item 2. does not apply. In addition,
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<math>A\left( 0 \right)=A\left( 1 \right)=0</math>, so the endpoints in item 3. cannot be maximum points (but rather the opposite, i.e. minimum points).
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We must therefore supposed that the maximum area is a critical point. We differentiate
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The function <math>A(x) = x(1-x^2)</math> is differentiable everywhere, so item 2 does not apply. In addition, <math>A(0) = A(1) = 0\,</math>, so the endpoints in item 3 cannot be maximum points (but rather the opposite, i.e. minimum points).
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<math>{A}'\left( x \right)=1\centerdot \left( 1-x^{2} \right)+x\centerdot \left( -2x \right)=1-3x^{2}</math>,
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We must therefore conclude that the maximum area is a critical point. We differentiate
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and the condition that the derivative should be zero gives that
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{{Displayed math||<math>A'(x) = 1\cdot (1-x^2) + x\cdot (-2x) = 1-3x^2\,,</math>}}
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<math>x=\pm {1}/{\sqrt{3}}\;</math>; however, it is only
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<math>x={1}/{\sqrt{3}}\;</math>
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which satisfies
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<math>0\le x\le 1</math>
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At the critical point, the second derivative
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<math>{A}''</math>
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has the value
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and the condition that the derivative should be zero gives that <math>x=\pm 1/\!\sqrt{3}</math>; however, it is only <math>x=1/\!\sqrt{3}</math> which satisfies <math>0\le x\le 1</math>.
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<math>{A}''\left( {1}/{\sqrt{3}}\; \right)=-6\centerdot \frac{1}{\sqrt{3}}<0</math>,
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At the critical point, the second derivative <math>A''(x)=-6x</math> has the value
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which shows that
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{{Displayed math||<math>A''\bigl( 1/\!\sqrt{3}\bigr) = -6\cdot\frac{1}{\sqrt{3}} < 0\,,</math>}}
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<math>{1}/{\sqrt{3}}\;</math>
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is a local maximum.
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The answer is that the point
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which shows that <math>x=1/\!\sqrt{3}</math> is a local maximum.
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<math>P</math>
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should be chosen so that
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The answer is that the point <math>P</math> should be chosen so that
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<math>P=\left( \frac{1}{\sqrt{3}} \right.,\left. 1-\left( \frac{1}{\sqrt{3}} \right)^{2} \right)=\left( \frac{1}{\sqrt{3}} \right.,\left. \frac{2}{3} \right)</math>.
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{{Displayed math||<math>P = \Bigl(\frac{1}{\sqrt{3}}, 1-\Bigl(\frac{1}{\sqrt{3}} \Bigr)^2\, \Bigr) = \Bigl(\frac{1}{\sqrt{3}}, \frac{2}{3} \Bigr)\,\textrm{.}</math>}}

Version vom 08:43, 21. Okt. 2008

If we call the x-coordinate of the point \displaystyle P \displaystyle x, then its y-coordinate is \displaystyle 1-x^{2}, because \displaystyle P lies on the curve \displaystyle y=1-x^{2}.

The area of the rectangle is then given by

\displaystyle A(x) = \text{(base)}\cdot\text{(height)} = x\cdot (1-x^2)

and we will try to choose \displaystyle x so that this area function is maximised.

To begin with, we note that, because \displaystyle P should lie in the first quadrant, \displaystyle x\ge 0 and also \displaystyle y=1-x^2\ge 0, i.e. \displaystyle x\le 1. We should therefore look for the maximum of \displaystyle A(x) when \displaystyle 0\le x\le 1\,.

There are three types of points which can maximise the area function:

  1. critical points,
  2. points where the function is not differentiable,
  3. endpoints of the region of definition.

The function \displaystyle A(x) = x(1-x^2) is differentiable everywhere, so item 2 does not apply. In addition, \displaystyle A(0) = A(1) = 0\,, so the endpoints in item 3 cannot be maximum points (but rather the opposite, i.e. minimum points).

We must therefore conclude that the maximum area is a critical point. We differentiate

\displaystyle A'(x) = 1\cdot (1-x^2) + x\cdot (-2x) = 1-3x^2\,,

and the condition that the derivative should be zero gives that \displaystyle x=\pm 1/\!\sqrt{3}; however, it is only \displaystyle x=1/\!\sqrt{3} which satisfies \displaystyle 0\le x\le 1.

At the critical point, the second derivative \displaystyle A''(x)=-6x has the value

\displaystyle A''\bigl( 1/\!\sqrt{3}\bigr) = -6\cdot\frac{1}{\sqrt{3}} < 0\,,

which shows that \displaystyle x=1/\!\sqrt{3} is a local maximum.

The answer is that the point \displaystyle P should be chosen so that

\displaystyle P = \Bigl(\frac{1}{\sqrt{3}}, 1-\Bigl(\frac{1}{\sqrt{3}} \Bigr)^2\, \Bigr) = \Bigl(\frac{1}{\sqrt{3}}, \frac{2}{3} \Bigr)\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:4“