1950 AHSME Problems/Problem 41
Contents
[hide]Problem
The least value of the function with is:
Solution
The vertex of a parabola is at for . Because , the vertex is a minimum. Therefore .
Solution 2
Using calculus, we find that the derivative of the function is , which has a critical point at . The second derivative of the function is ; since , this critical point is a minimum. As in Solution 1, plug this value into the function to obtain .
~ cxsmi
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 40 |
Followed by Problem 42 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.