Lösung 1.3:4
Aus Online Mathematik Brückenkurs 2
If we call the
FIGURE
The area of the rectangle is then given by
x
=
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).
