Difference between revisions of "2021 AMC 12A Problems/Problem 7"
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==Solution== | ==Solution== | ||
Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the trivial inequality(all squares are nonnegative) the minimum value for this is <math>\boxed{(D) 1}</math>, which can be achieved at <math>x=y=0</math>. ~aop2014 | Expanding, we get that the expression is <math>x^2+2xy+y^2+x^2y^2-2xy+1</math> or <math>x^2+y^2+x^2y^2+1</math>. By the trivial inequality(all squares are nonnegative) the minimum value for this is <math>\boxed{(D) 1}</math>, which can be achieved at <math>x=y=0</math>. ~aop2014 | ||
| + | |||
| + | ==Solution 2 (Beyond Overkill)== | ||
| + | Like solution 1, expand and simplify the original equation to <math>x^2+y^2+x^2y^2+1</math> and let <math>f(x, y) = x^2+y^2+x^2y^2+1</math>. To find local extrema, find where <math>\nabla f(x, y) = \boldsymbol{0}</math>. First, find the first with respect to x and y and find where they are <math>0</math>: | ||
| + | <cmath> \frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0</cmath> | ||
| + | <cmath> \frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0</cmath> | ||
| + | |||
| + | Thus, there is a local extreme at <math>(0, 0)</math>. Because this is the only extreme, we can assume that this is a minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning <math>f(0, 0)</math> is the minimum of <math>f(x, y)</math>. Plugging (0, 0) into f(x, y), we find 1 <math>\implies \boxed{\bold{(D)} \ 1}</math> | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== | ||
https://www.youtube.com/watch?v=P5al76DxyHY | https://www.youtube.com/watch?v=P5al76DxyHY | ||
| + | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=6|num-a=8}} | {{AMC12 box|year=2021|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:54, 11 February 2021
Contents
Problem
What is the least possible value of 
for real numbers 
and 
?
Solution
Expanding, we get that the expression is 
or 
. By the trivial inequality(all squares are nonnegative) the minimum value for this is 
, which can be achieved at 
. ~aop2014
Solution 2 (Beyond Overkill)
Like solution 1, expand and simplify the original equation to and let 
. To find local extrema, find where 
. First, find the first with respect to x and y and find where they are 
:
![\[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\]](http://latex.artofproblemsolving.com/2/c/8/2c8883627988c35283b34d3dfa17825ed8bf491d.png)
![\[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\]](http://latex.artofproblemsolving.com/2/f/d/2fde4a923b8182ddc56a7bc835d6d6b8a438e351.png)
Thus, there is a local extreme at 
. Because this is the only extreme, we can assume that this is a minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning 
is the minimum of 
. Plugging (0, 0) into f(x, y), we find 1 
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
