Posted by **bobby** on Tuesday, December 16, 2008 at 7:52pm.

Find two numbers whose sum is 10 for which the sum of their squares is a minimum.

- Calculus -
**Reiny**, Tuesday, December 16, 2008 at 7:55pm
two numbers : x and 10-x

S = x^2 + (10-x)^2

dS/dx = 2x - 2(10-x)

= 0 for max/min

x = 5

so the numbers are both 5

- Calculus -
**Damon**, Tuesday, December 16, 2008 at 7:56pm
x and (10-x)

s = x^2 + (10-x)^2

s = x^2 + 100 -20x + x^2

s = 2 x^2 - 20 x + 100

s/2 = x^2 -10 x + 100 we can minimize half the sum easier than the whole sum

That is a parabola and you could find the vertex but since you said this was "calculus" we will take the derivative and set to zero.

0 = 2 x - 10

x = 5

10-x = 5

- Calculus -
**Damon**, Tuesday, December 16, 2008 at 8:23pm
Interesting the answer is halfway between.

Exploring that

Say a sum of two numbers is s

We want to minimize the sum of squares of x^2 and (s-x)^2

sum = 2 x^2 -2sx

d sum/dx = 0 = 4 x -2s

x = s/2

so it works for any old sum, not just 10

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