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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$a,b,c,d,e \in R$

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

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

$find :\,\, e_{max}$

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

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

$find :\,\, e_{max}$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

$a,b,c,d,e \in R$

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

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

$find :\,\, e_{max}$

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

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

$find :\,\, e_{max}$

$a,b,c,d,e \in R$

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

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

$find :\,\, e_{max}$

When a=b=c=d=2 we get e=0 and when a=b=c=d=1.2 we get e=3.2.

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- Jan 25, 2013

- 1,225

$e_{max}=?$

and can you prove it ?

and can you prove it ?

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\(\displaystyle a=b=c=d\)

And so:

\(\displaystyle 4a+e=8\implies a=\frac{8-e}{4}\)

\(\displaystyle 4a^2+e^2=16\)

Substitute for $a$:

\(\displaystyle 4\left(\frac{8-e}{4} \right)^2+e^2=16\)

This simplifies to:

\(\displaystyle e(5e-16)=0\)

Hence:

\(\displaystyle e_{\max}=\frac{16}{5}\)

Yep. See MarkFL's post.$e_{max}=?$

and can you prove it ?