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

2 Answers
Jamen N.
Apr 4, 2018

#10+10 = 20#
#10^2 +10^2=200#.

Explanation:

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

Shwetank Mauria
Apr 4, 2018

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

Explanation:

If sum of two numbers is #20#,

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

Hence their sum of squares is

#x^2+(20-x)^2#

= #x^2+400-40x+x^2#

= #2x^2-40x+400#

= #2(x^2-20x+100-100)+400#

= #2(x-10)^2-200+400#

= #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#

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

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

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