Stewart §4.7 #51. Let a and b be positive numbers. Find the length of the shortest line segment that is cut off
by the first quadrant and passes through the point (a, b).
There are many different ways of describing a line that passes through (a, b): we could specify its x-intercept, its
y-intercept, its slope, etc. We’ll choose slope, though presumably a similar should work with any other description.
So, let y − b = m · (x − a) be a line through (a, b) with unknown slope m. We need to minimize the distance between
the x- and y-intercepts, so we first calculate that
b
m
= b − am.
0 − b = m · (xint − a) ⇒ xint = a −
yint − b = m · (0 − a) ⇒ yint
Now, note that it is equivalent to minimize the distance and to minimize the squared-distance. This is because
distances are always nonnegative, and the squaring function is increasing when restricted to the nonnegative numbers. This allows us to sidestep the annoying square-root sign in the distance formula when we take derivatives.
So, we first find the squared-distance (as a function of m) to be
2
s
2
2
b
b
2
2
+ (b − am)  = a −
+ (b − am) .
D=
a−
m
m
We want to minimize this with respect to m, so we find
dD
b
b
set
0 =
=2 a−
· 2 + 2 (b − am) · (−a)
dm
m
m
b
b
= 2· 2 a−
− 2a (b − am)
m
m
b
= 2 · 3 (am − b) + 2a(am − b)
m
b
= 2(am − b) ·
+
a
.
m3
(In going from the second line to the third we can indeed multiply the first summand by m/m since we’re clearly
already assuming m 6= 0, either because m already appears as a denominator or because m = 0 is obviously not
a solution to our original problem.) This implies that at least one of the factors is zero. If 2(am − b) = 0 then
m = b/a; this gives us the line through (0, 0) p
and (a, b), which is clearly not the solution to our original problem.
So (b/m3 ) + a = 0, which implies that m = 3 −b/a. This is indeed a minimum because our domain of interest
is m ∈ (−∞, 0) and as m approaches either end the distance approaches infinity. We can plug this back into our
squared-distance function to get

Dmin
2
r
=
=
(a + a1/3 b2/3 )2 + (b + a2/3 b1/3 )2
=
(a2 + 2a4/3 b2/3 + a2/3 b4/3 ) + (b2 + 2a2/3 b4/3 + a4/3 b2/3 )
=
a2 + 3a4/3 b2/3 + 3a2/3 b4/3 + b2
=
(a2/3 )3 + 3(a2/3 )2 (b2/3 ) + 3(a2/3 )(b2/3 )2 + (b2/3 )3
=
(a2/3 + b2/3 )3 .
b−a·
Thus the distance in question is (a2/3 + b2/3 )3/2 .
1
3
b
−
a
!2
a − qb  +
3
− ab