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Lösung 1.3:4

Aus Online Mathematik Brückenkurs 2

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If we call the x -coordinate of the point P

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

FIGURE

The area of the rectangle is then given by


Ax=  base *height =x1x2 .


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=1x20, 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=x1x2  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=11x2+x2x=13x2 ,

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 13  is a local maximum.

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


P=131132=1332 .