F.4 Math求解

let this two number are a and b

Then a+b=10

Want the min of a^2+b^2

Since b=10-a

We have

a^2+b^2
=a^2+(10-a)^2
=2a^2-20a+100
=2(a^2-10a+50)
=2[(a-5)^2+25]

So when a=b=5, the sum of their square is a minimum

a^2+b^2=5^2+5^2=50

let x +y =10
y = 10-x

s = x^2 +y^2
s = x^2 + (10-x)^2
s = x^2 + 100 - 20x +x^2
ds/dx = 4x -20 = 0
x=5

d^2s/dx^2 = 4 > 0 so will know this is the minimum
x=5
y=5

x^2 + y^2 = 50