Difference between revisions of "2021 AMC 12A Problems/Problem 7"

Revision as of 16:08, 11 February 2021

Problem

What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$ ?

$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution

Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the trivial inequality(all squares are nonnegative) the minimum value for this is $\boxed{(D) 1}$ , which can be achieved at $x=y=0$ . ~aop2014

Solution 2 (Beyond Overkill)

Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first with respect to x and y and find where they are $0$ : $\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0$ $\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0$

Thus, there is a local extreme at $(0, 0)$ . Because this is the only extreme, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{\bold{(D)} \ 1}$

~ DBlack2021

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

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.

@@ Line 12: / Line 12: @@
 <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 because the problem asks for the 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>
+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 because the problem asks for the 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 <math>(0, 0)</math> into <math>f(x, y)</math>, we find 1 <math>\implies \boxed{\bold{(D)} \ 1}</math>
 ~ DBlack2021

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