Find Min Value of $a^2+b^2+c^2+d^2+(a+b+c+d)^2$ w/ Given Constraint

[anemone]

Find the minimum of $a^2+b^2+c^2+d^2+(a+b+c+d)^2$ given $a+2b+3d+4d=1$ for all real $a,\,b,\,c$ and $d$.

 

[Albert1]

Find the minimum of $a^2+b^2+c^2+d^2+(a+b+c+d)^2$ given $a+2b+3d+4d=1$ for all real $a,\,b,\,c$ and $d$.

Spoiler

minimum=$\dfrac {1}{5}$

 

[anemone]

Spoiler

minimum=$\dfrac {1}{5}$

Hi Albert!

Spoiler

Nice try!:) But, I'm sorry, your answer isn't correct...:(

 

[Albert1]

Another method:

Spoiler

Using Cauchy Schwarz inequality:
$(a^2+b^2+c^2+d^2)\times (1^2+1^2+1^2+1^2)\ge (a+b+c+d)^2$
$\therefore (a^2+b^2+c^2+d^2)\ge \dfrac {(a+b+c+d)^2}{4}$
so we will find min$(\dfrac {5(a+b+c+d)^2}{4}$)
equality will occur when $a=b=c=d=\dfrac {1}{10}$ for $a+2b+3c+4d=1$
and we get minimum=$\dfrac {1}{5}$
what 's wrong with the answer ?

 

[anemone]

Another method:

Spoiler

Using Cauchy Schwarz inequality:
$(a^2+b^2+c^2+d^2)\times (1^2+1^2+1^2+1^2)\ge (a+b+c+d)^2$
$\therefore (a^2+b^2+c^2+d^2)\ge \dfrac {(a+b+c+d)^2}{4}$
so we will find min$(\dfrac {5(a+b+c+d)^2}{4}$)
equality will occur when $a=b=c=d=\dfrac {1}{10}$ for $a+2b+3c+4d=1$
and we get minimum=$\dfrac {1}{5}$
what 's wrong with the answer ?

Hi Albert,

Spoiler

At first glance, there seems like your solution works...but looking at it more closely, I think you have transformed the given target expression and what you are after now is the minimum of $\dfrac {5(a+b+c+d)^2}{4}$, and upon checking it with wolfram, $\dfrac {5(a+b+c+d)^2}{4}\ge 0$ if $a+2b+3c+4d=1$...

http://www.wolframalpha.com/input/?i=Find+the+minimum+of+5%28a%2Bb%2Bc%2Bd%29^2%2F4+if+a%2B2b%2B3c%2B4d%3D1

 

[MarkFL]

My solution:

Spoiler

If we set the objective function as:

\(\displaystyle f(a,b,c,d)=a^2+b^2+c^2+d^2+(a+b+c+d)^2\)

subject to the constraint:

\(\displaystyle g(a,b,c,d)=a+2b+3d+4d-1=0\)

Then, Lagrange multipliers gives us:

\(\displaystyle 2a+2(a+b+c+d)=\lambda\)

\(\displaystyle 2b+2(a+b+c+d)=2\lambda\)

\(\displaystyle 2c+2(a+b+c+d)=3\lambda\)

\(\displaystyle 2d+2(a+b+c+d)=4\lambda\)

Now, adding these 4 equations, we find:

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

and then we observe that the second equation gives us $b=0$ and we are left with:

\(\displaystyle 3a+c+d=0\)

\(\displaystyle c-a-d=0\)

\(\displaystyle a+c=0\)

Now, these imply:

\(\displaystyle c=-a\)

\(\displaystyle d=-2a\)

And so substituting into the constraint, we find:

\(\displaystyle a+2(0)+3(-a)+4(-2a)=1\implies a=-\frac{1}{10}\)

and so we have the critical point:

\(\displaystyle (a,b,c,d)=\left(-\frac{1}{10},0,\frac{1}{10},\frac{1}{5}\right)\)

And we then find:

\(\displaystyle f\left(-\frac{1}{10},0,\frac{1}{10},\frac{1}{5}\right)=\frac{1}{100}+0+\frac{1}{100}+\frac{1}{25}+\frac{1}{25}=\frac{1}{10}\)

Now, to determine whether this extremum is a minimum or maximum, let's look at the point:

\(\displaystyle (a,b,c,d)=(1,0,0,0)\)

We find:

\(\displaystyle f(1,0,0,0)=1+1=2>\frac{1}{10}\)

Hence, we may then conclude:

\(\displaystyle f_{\min}=\frac{1}{10}\)

 

[anemone]

Very well done, MarkFL!

I hope Albert or anyone else could try again with the Cauchy Schwartz inequality, it can still be used in a more implicit way so to find the minimum that agrees with Mark's answer.

 

[Albert1]

Very well done, MarkFL!

I hope Albert or anyone else could try again with the Cauchy Schwartz inequality, it can still be used in a more implicit way so to find the minimum that agrees with Mark's answer.

please show your solution using the Cauchy Schwartz inequality

 

[anemone]

Here goes:

Find the minimum of $a^2+b^2+c^2+d^2+(a+b+c+d)^2$ given $a+2b+3d+4d=1$ for all real $a,\,b,\,c$ and $d$.

I didn't realize until now that I had a typo in the original problem, where the $3d$ in $a+2b+3d+4d=1$ should be a $3c$. Sorry! (Tmi)

Solution of other:

Spoiler

$((-a)+(0b)+(c)+(2d)+2(a+b+c+d))^2\le ((-1)^2+(0)^2+1^2+2^2+2^2)(a^2+b^2+c^2+d^2+(a+b+c+d)^2)$

$(a+2b+3d+4d)^2\le (1+0+1+4+4)(a^2+b^2+c^2+d^2+(a+b+c+d)^2)$

$1\le 10(a^2+b^2+c^2+d^2+(a+b+c+d)^2)$

$a^2+b^2+c^2+d^2+(a+b+c+d)^2 \ge \dfrac{1}{10}$, equality occurs when $a=-0.1,\,b=0,\,c=0.1,\,d=0.2$.

 

[Albert1]

Here goes:
I didn't realize until now that I had a typo in the original problem, where the $3d$ in $a+2b+3d+4d=1$ should be a $3c$. Sorry! (Tmi)

Solution of other:

Spoiler

$((-a)+(0b)+(c)+(2d)+2(a+b+c+d))^2\le ((-1)^2+(0)^2+1^2+2^2+2^2)(a^2+b^2+c^2+d^2+(a+b+c+d)^2)$

$(a+2b+3d+4d)^2\le (1+0+1+4+4)(a^2+b^2+c^2+d^2+(a+b+c+d)^2)$

$1\le 10(a^2+b^2+c^2+d^2+(a+b+c+d)^2)$

$a^2+b^2+c^2+d^2+(a+b+c+d)^2 \ge \dfrac{1}{10}$, equality occurs when $a=-0.1,\,b=0,\,c=0.1,\,d=0.2$.

very skillful !