Lösung 1.3:4

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If we call the
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<center> [[Image:1_3_4-1(3).gif]] </center>
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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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<center> [[Image:1_3_4-2(3).gif]] </center>
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<math>x</math>, then its
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<math>y</math>
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<center> [[Image:1_3_4-3(3).gif]] </center>
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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
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<math>A\left( x \right)=</math>
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base *height
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<math>=x\centerdot \left( 1-x^{2} \right)</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:
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1. critical points,
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2. points where the function is not differentiable,
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3. 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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<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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and the condition that the derivative should be zero gives that
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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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<math>{A}''\left( {1}/{\sqrt{3}}\; \right)=-6\centerdot \frac{1}{\sqrt{3}}<0</math>,
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which shows that
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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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<math>P</math>
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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>.

Version vom 09:48, 16. Okt. 2008

If we call the \displaystyle x -coordinate of the point \displaystyle P

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

FIGURE

The area of the rectangle is then given by


\displaystyle A\left( x \right)= base *height \displaystyle =x\centerdot \left( 1-x^{2} \right).


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

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


\displaystyle {A}'\left( x \right)=1\centerdot \left( 1-x^{2} \right)+x\centerdot \left( -2x \right)=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}'' has the value


\displaystyle {A}''\left( {1}/{\sqrt{3}}\; \right)=-6\centerdot \frac{1}{\sqrt{3}}<0,

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

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


\displaystyle 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).

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:4“