MHB S8.3.7.4. The sum of two positive numbers is 16.

[karush]

3.7.4. The sum of two positive numbers is 16.

What is the smallest possible value of the sum of their squares?

$x+y=16\implies y=16-x$
Then
$x^2+(16-x)^2=2 x^2 - 32x + 256$

So far
... Hopefully

 

[skeeter]

keep going ... ATQ

 

[karush]

[3.7.4. The sum of two positive numbers is 16.
What is the smallest possible value of the sum of their squares?

$x+y=16\implies y=16-x$
Then
$z=x^2+(16-x)^2=2 x^2 - 32x + 256$
Then
$z'=4x-32$
So
$z'(x)=0\quad x=8$
Check
8^2+8^2=128

 

[topsquark]

[3.7.4. The sum of two positive numbers is 16.
What is the smallest possible value of the sum of their squares?

$x+y=16\implies y=16-x$
Then
$z=x^2+(16-x)^2=2 x^2 - 32x + 256$
Then
$z'=4x-32$
So
$z'(x)=0\quad x=8$
Check
8^2+8^2=128

Question: Is that the smallest value for x?

-Dan

 

[Prove It]

Question: Is that the smallest value for x?

-Dan

Question: Is that what was asked in the question? I thought they were asking for the smallest sum of squares...

 

[topsquark]

Question: Is that what was asked in the question? I thought they were asking for the smallest sum of squares...

Okay. Yes, you are right. I was trying to get karush to prove that this was the minimum answer in order to round out his solution. He never proved that.

-Dan

 

[HallsofIvy]

Rather than use something as "advanced" as setting the derivative to 0, I would "complete the square". [math]2x^2- 32x+ 256= 2(x^2- 16x+ 128)= 2(x^2- 16x+ 64+ 64)= 2((x- 8)^2+ 64)[/math].

That will be smallest when x= 8 and then the value will be 2(64)= 128.