Position the phone in the first quadrant so that two of its corners are at (a,0) and (0,a) and the line segment from (a,0) to (0,a) is one of the 3.5" sides. Now place the pencil along the opposite side of the (a,0), (0,a) side, such that it intersects the axes at (b,0) and (0,b). Call the pencil side the "outer" side and the opposite side the "inner" side.
Now imagine a line segment from O (the origin) to the outer side, intersecting the outer side at a 90 degree angle at a point P. Clearly it also intersects the inner side at a 90 degree angle as well at some other point Q. Note that the segment from the origin to P is the same length as the segment from P to (b,0) because the triangle defined by those three points is an isosceles right triangle.
We can find a as follows:
a**2 + a**2 = 3.5**2
a = sqrt(3.5**2/2)
Now we can find the length of segment OQ as follows by considering the triangle formed by O, Q, and (a,0). Note that the segment from Q to (a,0) is of length 3.5/2.
(OQ)**2 + (3.5/2)**2 = sqrt( 3.5**2/2 )**2 (that is, a**2 where a is given above)
(OQ)**2 + 3.5**2/4 = 3.5**2/2
(OQ)**2 = 3.5**2/4
OQ = 3.5/2
We know that the length of QP is 2 (since it's parallel to one of the 2-inch sides).
So OP = OQ + QP = 3.5/2 + 2 = 7.5/2
And since that equals half the length of the pencil, the pencil is 7.5" long.
I believe that the shortest hypotenuse for a right triangle is when the legs are equal, so this would be the optimal case, as other positions of the phone would make the legs of the triangle formed by the axes and pencil unequal. The case where the phone's sides run along the axes is a degenerate case of this positioning, and is suboptimal at any rate.
So: How did the above authors get 7.5" so easily? Did I miss a much simpler way of working it out?