The sum of two numbers is 20. Find the minimum possible sum of their squares?

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The sum of two numbers is 20. Find the minimum possible sum of their squares?

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Answer 1 · 2018-04-04T03:06:17

$10 + 10 = 20$ $10+10 = 20$
$10^{2} + 10^{2} = 200$ $10^2 +10^2=200$ .

Explanation:

$a + b = 20$ $a+b=20$
$a^{2} + b^{2} = x$ $a^2 + b^2= x$
For $a$ $a$ and $b$ $b$ :
$1^{2} + 19^{2} = 362$ $1^2+19^2=362$
$2^{2} + 18^{2} = 328$ $2^2+18^2=328$
$3^{2} + 17^{2} = 298$ $3^2+17^2=298$
From this, you can see that closer values of $a$ $a$ and $b$ $b$ will have a smaller sum. Thus, for $a = b$ $a=b$ , $10 + 10 = 20$ $10+10 = 20$ and $10^{2} + 10^{2} = 200$ $10^2 +10^2=200$ .

Answer 2 · 2018-04-04T03:06:43

Minimum value of sum of squares of two numbers is $200$ $200$ , which is when both numbers are $10$ $10$

Explanation:

If sum of two numbers is $20$ $20$ ,

let one number be $x$ $x$ and then other number would be $20 - x$ $20-x$

Hence their sum of squares is

$x^{2} + {(20 - x)}^{2}$ $x^2+(20-x)^2$

= $x^{2} + 400 - 40 x + x^{2}$ $x^2+400-40x+x^2$

= $2 x^{2} - 40 x + 400$ $2x^2-40x+400$

= $2 (x^{2} - 20 x + 100 - 100) + 400$ $2(x^2-20x+100-100)+400$

= $2 {(x - 10)}^{2} - 200 + 400$ $2(x-10)^2-200+400$

= $2 {(x - 10)}^{2} + 200$ $2(x-10)^2+200$

Observe that the sum of squares of two numbers is sum of two positive numbers, one of whom is a constant i.e. $200$ $200$

and other $2 {(x - 10)}^{2}$ $2(x-10)^2$ , which can change according to value of $x$ $x$ and its least value could be $0$ $0$ , when $x = 10$ $x=10$

Hence minimum value of sum of squares of two numbers is $0 + 200 = 200$ $0+200=200$ , which is when $x = 10$ $x=10$ , which is when both numbers are $10$ $10$ .

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The sum of two numbers is 20. Find the minimum possible sum of their squares?

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