Difference between revisions of "1978 USAMO Problems/Problem 1"
Mathlobster (talk | contribs) (→Solution 2) |
Mathlobster (talk | contribs) (→Solution 2) |
||
Line 18: | Line 18: | ||
Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem. | Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem. | ||
We get the following equations: | We get the following equations: | ||
+ | |||
<math>\newline(1) a+b+c+d+e=8\newline | <math>\newline(1) a+b+c+d+e=8\newline | ||
(2) a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\newline | (2) a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\newline | ||
Line 24: | Line 25: | ||
(5) 0=\lambda+2c\mu\newline | (5) 0=\lambda+2c\mu\newline | ||
(6) 0=\lambda+2d\mu\newline | (6) 0=\lambda+2d\mu\newline | ||
− | (7) 1=\lambda+2e\mu | + | (7) 1=\lambda+2e\mu</math> |
+ | |||
If <math>\mu=0</math>, then <math>\lambda=0</math> according to <math>(6)</math> and <math>\lambda=1</math> according to <math>(7)</math>, so <math>\mu \neq 0</math>. Setting the right sides of <math>(3)</math> and <math>(4)</math> equal yields <math>\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b</math>. Similar steps yield that <math>a=b=c=d</math>. Thus, <math>(1)</math> becomes <math>4d+e=8</math> and <math>(2)</math> becomes <math>4d^{2}+e^{2}=16</math>. Solving the system yields <math>e=0,\frac{16}{5}</math>, so that maximum possible value of <math>e</math> is <math>\frac{16}{5}</math> | If <math>\mu=0</math>, then <math>\lambda=0</math> according to <math>(6)</math> and <math>\lambda=1</math> according to <math>(7)</math>, so <math>\mu \neq 0</math>. Setting the right sides of <math>(3)</math> and <math>(4)</math> equal yields <math>\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b</math>. Similar steps yield that <math>a=b=c=d</math>. Thus, <math>(1)</math> becomes <math>4d+e=8</math> and <math>(2)</math> becomes <math>4d^{2}+e^{2}=16</math>. Solving the system yields <math>e=0,\frac{16}{5}</math>, so that maximum possible value of <math>e</math> is <math>\frac{16}{5}</math> | ||
Revision as of 12:55, 30 April 2016
Contents
Problem
Given that are real numbers such that
,
.
Determine the maximum value of .
Solution 1
Accordting to Cauchy-Schwarz Inequalities, we can see
thus,
Finally,
that mean,
so the maximum value of
is
from: Image from Gon Mathcenter.net
Solution 2
Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem. We get the following equations:
If , then
according to
and
according to
, so
. Setting the right sides of
and
equal yields
. Similar steps yield that
. Thus,
becomes
and
becomes
. Solving the system yields
, so that maximum possible value of
is
See Also
1978 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.