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

288

1156

366

576

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.