Problem for March 2004

Communicated by Dan Jurca

Solution by Kurt Luoto

Suppose a Î R, b Î R, a < b, f:[a,b]® R, and

x Î [a,b] Þ 0 < f″(x).

Find all values c Î [a,b] such that the area of the region bounded by

· the graph of y = f(x)

· the line tangent to the graph at the point (c,f(c))

· the lines x = a and x = b

is minimum. (The region for typical c is the shaded area in the figure below.)

______________________________________________

We first note the general facts that given a real-valued function f(x) and real constant c, then the value x minimizes f(x) iff x minimizes f(x)+c. If c > 0 then x minimizes f(x) iff x minimizes c∙f(x), while if c < 0 then x minimizes f(x) iff x maximizes c∙f(x).

Let l_c(x) denote the line tangent to the graph at the point (c,f(c)). Since 0 < f″(x) on the interval, the graph of y = f(x) is convex and therefore the graph of y = l_c(x) lies below the graph of y = f(x). Then it is clear that the problem asks us to find values of c that minimize . Since is a constant, this is the same as finding values of c that maximize .
Now l_c(x) = f(c) + f′(c)(x - c), so

.

Now letting c vary in this last expression and renaming c as x, we need to find values of x Î [a,b] that maximize

.

Letting m = (a + b)/2, i.e. the midpoint of [a,b], and noting that b-a > 0, this means we are seeking to maximize the function g(x) = (m-x) f′(x)+ f(x) on the interval [a,b]. Taking the derivative, g′(x) = (m-x) f″(x) - f′(x) + f′(x) = (m-x) f″(x), since 0 < f″(x) on the interval, we see that g′(x) > 0 for x < m, g′(x) = 0 for x = m, and g′(x) < 0 for x > m. Thus g(x) has a global maximum on [a,b] at x = m, and so the region in the problem statement is minimized at c = m = (a + b)/2, the midpoint of [a,b], and only at m.