121616949-math.128 - v w a and b then it is easy to...

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114 Chapter 6 Applications of the Derivative We want to find the minimum value of f when x is between 0 and a . As usual we set f ( x ) = 0 and solve for x : 0 = f ( x ) = - 1 v + x w x 2 + b 2 w p x 2 + b 2 = vx w 2 ( x 2 + b 2 ) = v 2 x 2 w 2 b 2 = ( v 2 - w 2 ) x 2 x = wb v 2 - w 2 Notice that a does not appear in the last expression, but a is not irrelevant, since we are interested only in critical values that are in [0 , a ], and wb/ v 2 - w 2 is either in this interval or not. If it is, we can use the second derivative to test it: f ( x ) = b 2 ( x 2 + b 2 ) 3 / 2 w . Since this is always positive there is a local minimum at the critical point, and so it is a global minimum as well. If the critical value is not in [0 , a ] it is larger than a . In this case the minimum must occur at one of the endpoints. We can compute f (0) = a v + b w f ( a ) = a 2 + b 2 w but it is difficult to determine which of these is smaller by direct comparison. If, as is likely in practice, we know the values of
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Unformatted text preview: v , w , a , and b , then it is easy to determine this. With a little cleverness, however, we can determine the minimum in general. We have seen that f ±± ( x ) is always positive, so the derivative f ± ( x ) is always increasing. We know that at wb/ √ v 2-w 2 the derivative is zero, so for values of x less than that critical value, the derivative is negative. This means that f (0) > f ( a ), so the minimum occurs when x = a . So the upshot is this: If you start farther away from C than wb/ √ v 2-w 2 then you always want to cut across the sand when you are a distance wb/ √ v 2-w 2 from point C . If you start closer than this to C , you should cut directly across the sand....
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