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Lösung 1.3:4

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.3:4 moved to Solution 1.3:4: Robot: moved page)
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{{NAVCONTENT_START}}
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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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{{NAVCONTENT_STOP}}
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-coordinate of the point
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{{NAVCONTENT_START}}
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<math>P</math>
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<center> [[Image:1_3_4-2(3).gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>x</math>, then its
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{{NAVCONTENT_START}}
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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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{{NAVCONTENT_STOP}}
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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
 +
<math>P</math>
 +
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
 +
<math>P</math>
 +
should be chosen so that
 +
 
 +
 
 +
<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 x -coordinate of the point P

x, then its y -coordinate is 1−x2, because P lies on the curve y=1−x2.

FIGURE

The area of the rectangle is then given by


Ax=  base *height =x1−x2 .


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

To begin with, we note that, because P should lie in the first quadrant, x0 and also y=1−x20, i.e. x1. We should therefore look for the maximum of Ax  when 0x1.

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 Ax=x1−x2  is differentiable everywhere, so item 2. does not apply. In addition, A0=A1=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


Ax=11−x2+x−2x=1−3x2 ,

and the condition that the derivative should be zero gives that x=13 ; however, it is only x=13  which satisfies 0x1 At the critical point, the second derivative A has the value


A13=−6130 ,

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.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:4“