Difference between revisions of "2019 AMC 10A Problems/Problem 19"
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<math>2y+10=0</math> | <math>2y+10=0</math> | ||
− | <math>2y | + | <math>2y=-5</math> |
<math>y=-5,0</math> | <math>y=-5,0</math> | ||
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''Note'': We could also have used the result that minimum/maximum point of a parabola <math>y = ax^2 + bx + c</math> occurs at <math>x=-\frac{b}{2a}</math>. | ''Note'': We could also have used the result that minimum/maximum point of a parabola <math>y = ax^2 + bx + c</math> occurs at <math>x=-\frac{b}{2a}</math>. | ||
+ | |||
+ | ''Note 2:'' This solution is somewhat "lucky", since when we define variables to equal a function, and create another function out of these variables, the domain of such function may vary from the initial one. This is important because the maximum and minimum value of a function is dependent on its domain, e.g: | ||
+ | |||
+ | <math>f(x)=x^2</math> has no maximum value in the the integers, but once restricting the domain to <math>(-5, 5)</math> the maxumum value of <math>f(x)</math> is <math>25</math>. | ||
+ | |||
+ | Also, observe that if we were to evaluate this by taking the derivative of <math>(x+1)(x+2)(x+3)(x+4)+2019</math>, we would get <math>-5</math> as the <math>x</math>-value to obtain the minimum <math>y</math>-value of this expression. It can be seen that <math>-5</math> is actually an inflection point, instead of a minimum or maximum. | ||
+ | |||
+ | -Benedict T (countmath1) | ||
==Solution 4(guess with answer choices)== | ==Solution 4(guess with answer choices)== |
Revision as of 18:14, 23 June 2022
Contents
Problem
What is the least possible value of where is a real number?
Solution 1
Grouping the first and last terms and two middle terms gives , which can be simplified to . Noting that squares are nonnegative, and verifying that for some real , the answer is .
Solution 2
Let . Then the expression becomes .
We can now use the difference of two 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 .
Solution 3 (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 .
Note: We could also have used the result that minimum/maximum point of a parabola occurs at .
Note 2: This solution is somewhat "lucky", since when we define variables to equal a function, and create another function out of these variables, the domain of such function may vary from the initial one. This is important because the maximum and minimum value of a function is dependent on its domain, e.g:
has no maximum value in the the integers, but once restricting the domain to the maxumum value of is .
Also, observe that if we were to evaluate this by taking the derivative of , we would get as the -value to obtain the minimum -value of this expression. It can be seen that is actually an inflection point, instead of a minimum or maximum.
-Benedict T (countmath1)
Solution 4(guess with answer choices)
The expression is negative when an odd number of the factors are negative. This happens when or . Plugging in or yields , which is very close to . Thus the answer is .
Solution 5 (using the answer choices)
Answer choices , , and are impossible, since can be negative (as seen when e.g. ). Plug in to see that it becomes , so round this to .
We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.
Solution 6 (no words)
.
.
.
Solution 7(naive solution)
Since we can obviously have the first part of the equation to be negative, let be -. Calculating, we find that it is a little more than 2018, and since we're given the choice of , we guess that it is .
Video Solutions
https://www.youtube.com/watch?v=Lis8yKT9WXc (less than 2 minutes)
- https://youtu.be/NRa3VnjNVbw - Education, the Study of Everything
- https://www.youtube.com/watch?v=Mfa7j2BoNjI
- https://youtu.be/tIzJtgJbHGc - savannahsolver
- https://youtu.be/3dfbWzOfJAI?t=3319 - pi_is_3.14
- https://youtu.be/GmUWIXXf_uk?t=1134 ~ pi_is_3.14
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.