Question

Two numbers have a difference of 24. What is the sum of their squares if it is a minimum?

Answer 1

Let "a" and "b" be some number where:

a - b = 24

We want to find where a^2 + b^2 is a minimum.  Instead of just logically figuring out that the answer is where a=b=12, I'll just use derivatives.

So we can first substitute for "a" where a = b+24

So we have (b+24)^2 + b^2 = b^2 +48b +576 + b^2
And that equals 2b^2 +48b +576

Then we take the derivative and set it equal to zero:

4b +48 = 0
4(b+12) = 0
b + 12 = 0
b = -12

Thus "a" must equal 12.

So:
a = 12
b = -12

And the sum of those two numbers squared is (12)^2 + (-12)^2 = 144 + 144 = 288.

The smallest sum is 288.