please help me with the following calculus optimization problem?

a+b= 25
We have (a+b)^2<=2(a^2+b^2)
=>(a^2+b^2)>=((a+b)^2)/2 = (25^2)/2 = constant
 Left side is equal to right side when and only when a=b
=>a=b=12.5

Let the numbers be a and b.

a + b = 25
a² + b² = minimum

b = 25 - a
=> a² + (25 - a)²
=> 2a² - 50a + 625

Let f = f(a)

f(a) = 2a² - 50a + 625
=> df/da = 4a - 50
=> 4a - 50 = 0
=> a = 12.5

You did not specify the number had to be integers, so this is a valid result.

Then you find b ----- b = 25 - a = 12.5

=> a = b = 12.5 which makes sense due to symmetry.

X + Z = 25   i.e. sum of X & Z = 25

so Z = 25 - X

also

Y = X^2 + Z^2  where Y is the value that you want to find the minimum for

substitute Z = 25 - X  into the above equation, which gives

Y = X^2 + (25 - X)^2

multiply out the bracket

Y = X^2 + X^2 -50X + 625   collect like terms

Y = 2X^2 - 50X + 625

differentiate the above with respect to X to find the gradient

dY/dX = 4X - 50

for a maximum or minimum the gradient = zero

i.e. 0 = 4X - 50

i.e. 4X = 50

so X = 50/4 = 12.5

you could check if this is a maximum or minimum value by drawing a graph

but using calculus is easiest

differentiate Y = 4X - 50 with respect to X to get

d2Y/dX^2 = 4

as this value is a positive value it tells you that this is a minimum.

So

X = 12.5 for a minimum value for the sum of squares

so what is the other number

12.5 + Z = 25, so Z also = 12.5

Not sure if this is correct because I didn't get whole numbers but I did get the smallest sum of the squares.

y = x^2 + (25 - x)^2
y = x^2 -50x +625 + x^2
y = 2x^2 -50x +625

y' = 4x - 50

0 = 4x - 50
50 = 4x
x = 12.5

12.5 + y = 25
y = 12.5

12.5^2 = 156.25

156.25 + 156.25 = 312.5


If your looking for integers, then I'd go with 12 and 13. The sum of their squares is 313.