Lösung 1.3:5

Aus Online Mathematik Brückenkurs 2

The channel holds most water when its cross-sectional area is greatest.

By dividing up the channel's cross-section into a rectangle and two triangles we can, with the help of a little trigonometry, determine the lengths of the rectangle's and triangles' sides, and thence the cross-sectional area.

The area of the cross-section is

A()=1010cos+22110cos10sin=100cos(1+sin).

If we limit the angle to lie between 0 and 2, the problem can be formulated as:

Maximise the function A()=100cos(1+sin) when

02.

The area function is a differentiable function and the area is least when =0 or =2, so the area must assume its maximum at a critical point of the area function.

We differentiate the area function:

A()=100(−sin)(1+sin)+100coscos=−100sin−100sin2+100cos2.

At a critical point A()=0 and this gives us the equation

sin+sin2−cos2=0

after eliminating the factor -100. We replace cos2 with 1−sin2 (according to the Pythagorean identity) to obtain an equation solely in sin,

sin+sin2−(1−sin2)2sin2+sin−1=0=0.

This is a second-degree equation in sin and completing the square gives that

2sin+412−2412−1sin+412=0=916

and we obtain \displaystyle \sin\alpha = -\tfrac{1}{4}\pm \tfrac{3}{4}, i.e. \displaystyle \sin \alpha = -1 or \displaystyle \sin \alpha = \tfrac{1}{2}\,.

The case \displaystyle \sin \alpha =-1 is not satisfied for \displaystyle 0\le \alpha \le \pi/2 and \displaystyle \sin \alpha = \tfrac{1}{2} gives \displaystyle \alpha = \pi/6. Thus \displaystyle \alpha = \pi/6\, is a critical point.

If we summarize, we know therefore that the cross-sectional area has local minimum points at \displaystyle \alpha = 0 and \displaystyle \alpha = \pi/2 and that we have a critical point at \displaystyle \alpha = \pi/6\,. This critical point must be a maximum, which we can also show using the second derivative,

which is negative at \displaystyle \alpha = \pi/6,

There are no local maximum points other than \displaystyle \alpha = \pi/6\,, which must therefore also be a global maximum.