We are not told that the solution requires integer values so we need to show that this is true. This requires some straightforward calculus.

Let \(x\) be the distance along the river East of Chuck’s ranch.  And let \(c\) and \(d\) be the direct distances from the point \(x\) on the river to Chuck’s and Dwayne’s ranches respectively.  We hence have:

\[c =\sqrt{(x)^2 + 16^2} \qquad  d=\sqrt{(30-x)^2+56^2}\]

For Dwayne’s desired solution of equal ranch/bridge distances we can simply equate \(c^2\) and \(d^2\) and solve the resulting equation for \(x\):  \[x^2  + 16^2 = (30-x)^2 + 56^2\  \quad \Rightarrow \quad 60 x = 900+56^2-16^2 \quad \Rightarrow \quad x=63\]

For Chuck’s desired solution we have to maximise the difference between the two ranch/bridge distances by finding the derivative: \(\frac{d}{dx}(c-d)\) and setting this to zero.  This is equivalent to \(\frac{d}{dx}(c) = \frac{d}{dx}(d)\) which will be the form used here:  \[\frac{x}{\sqrt{x^2+16^2}} = \frac{x-30}{\sqrt{(30-x)^2+56^2}}\] If we now cross multiply and simplify the result we obtain: \[960(3x^2 + 16x-240) = 0 \quad \Rightarrow \quad 960(3x-20)(x + 12) = 0\] This gives two values:  \(x=20/3\) and \(x=-12\) for which the latter integer result can be shown to provide the maximum difference by substitution into the expressions for \(c\, – \,d\).

While we now have a complete solution, knowing that only integer solutions are involved means that we can start at Chuck’s westward position on the river and look for integer distances left and right of this point on the river where both ranch/bridge distances are integers.