Congratulations to this week's winner

Kevin Bourrillion

A correct solution was also received from Yan Fridman.

One can solve the problem as a straightforward minimization problem in differential calculus. But Kevin Bourrillion's sparkling solution requires no knowledge of calculus at all. Here 'tis.

First note that a simple vertical shift of the graph allows one to assume that the graph of f(x) is positive throughout the interval in question. Also the condition on the concavity means that the graph of any tangent line is above the graph of f(x) throughout, except, of course, at the point of tangency, where they are equal. Next, note that since the area under the graph of f(x) is a constant, it's enough to minimize the area under the graph of T_c(x), the tangent line at the point c.

This area is that of a trapezoid and as such is given by the formula

(b - a) (T_c(a) + T_c(b))/2. Since the tangent is a staight line, this is further equal to (b - a) T_c((a + b)/2). By the observation above, the term T_c((a + b)/2) is greater than f((a + b)/2) for all values of c not equal to (a+b)/2. Therefore, the area is minimized when c = (a + b)/2, that is, for the tangent line at the midpoint.

You are visitor number 4139 to this page.
Page last updated 24 November 1997.