Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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(inspired by solution by oO8715_alexOo) | (inspired by solution by oO8715_alexOo) | ||
| − | Note: The minimum/maximum of a parabola occurs at | + | Note: The minimum/maximum of a parabola occurs at <math>x=-b/2a</math>. |
==See Also== | ==See Also== | ||
Revision as of 01:28, 10 February 2019
Problem
What is the least possible value of
![\[(x+1)(x+2)(x+3)(x+4)+2019\]](http://latex.artofproblemsolving.com/1/2/8/128007ee3b97f3ff741a1e668139edf8c13ba7e1.png)
where 
is a real number?
Solution
Grouping the first and last terms and two middle terms gives 
which can be simplified as 
. Since squares are nonnegative, the answer is 
Solution 2
Let 
. Then 
becomes 
We can use difference of squares to get 
, and expand this to get 
.
Refactor this by completing the square to get 
, which has a minimum value of 
. The answer is thus 
-WannabeCharmander
Solution 3 (using calculus)
Similar to Solution 1, grouping the first and last terms and the middle terms, we get .
Letting 
, we get the expression 
. Now, we can find the critical points of 
to minimize the function.
To minimize the result, we use 
. Hence, the minimum is 
, so 
(inspired by solution by oO8715_alexOo)
Note: The minimum/maximum of a parabola occurs at 
.
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
