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Difference between revisions of "1978 USAMO Problems/Problem 1"

(→‎Solution 2)
(→‎Solution 2)
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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\newline</math>
+
(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 11:55, 30 April 2016

Problem

Given that $a,b,c,d,e$ are real numbers such that

$a+b+c+d+e=8$,

$a^2+b^2+c^2+d^2+e^2=16$.

Determine the maximum value of $e$.

Solution 1

Accordting to Cauchy-Schwarz Inequalities, we can see $(1+1+1+1)(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2$ thus, $4(16-e^2)\geqslant (8-e)^2$ Finally, $e(5e-16) \geqslant 0$ that mean, $\frac{16}{5} \geqslant e \geqslant 0$ so the maximum value of $e$ is $\frac{16}{5}$

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:

$\newline(1) a+b+c+d+e=8\newline (2) a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\newline (3) 0=\lambda+2a\mu\newline (4) 0=\lambda+2b\mu\newline (5) 0=\lambda+2c\mu\newline (6) 0=\lambda+2d\mu\newline (7) 1=\lambda+2e\mu$

If $\mu=0$, then $\lambda=0$ according to $(6)$ and $\lambda=1$ according to $(7)$, so $\mu \neq 0$. Setting the right sides of $(3)$ and $(4)$ equal yields $\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b$. Similar steps yield that $a=b=c=d$. Thus, $(1)$ becomes $4d+e=8$ and $(2)$ becomes $4d^{2}+e^{2}=16$. Solving the system yields $e=0,\frac{16}{5}$, so that maximum possible value of $e$ is $\frac{16}{5}$

See Also

1978 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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