Question

Optimization question

The sum of two positive numbers is 16. What is the smallest possible value of the sum of their squares?

Answer 1

uhhh 128?

Answer 2

#'s are x and (16-x). Equation: y = x^2 + (16 - x)^2. Start by taking first derivative.

Answer 3

first derivative of each?

Answer 4

let's say a+a=16
now \[a^2+a^2=?\] 2a^2= y \[2a^2=y\] now \[\frac{ dy }{ dx }=2*2a=4a \] and \[\frac{ d ^{2}y }{ dx^2 }=4\] so minimum value is attained at a=0 therefore sum of their squares =0

Answer 5

woah. one by one @jasonxx

Answer 6

:)

Answer 7

flashing whole solutions just confuse people...better to discuss the steps one by one...

anyway...derivative of what @tcarroll010 ?

Answer 8

the 2 numbers need not be the same, so continuing, if y = x^2 + (16 - x)^2, then y' = 2x - 2(16 - x) = 4x - 32 = 4(x-8)

Answer 9

ironic how i just said flashing the whole solutions confuse people.....

Answer 10

At x = 8 the first derivative is = 0 and it goes from negative to positive, so at x = 8, we have a minimum. Could also have taken the second derivative and see that at this point, the curve is "concave up" and hence a minimum.

Answer 11

hmm so y' is 4(x-8)...i suppose this is equated to 0?

Answer 12

so x = 8

Answer 13

Yes. At 8 it's 0. The first derivative

Answer 14

so if x = 8..then y = 8 too...

then the sum of squares will be 128?

Answer 15

Yes.

Answer 16

thanks

Answer 17

You're quite welcome!