Tricky question
The sum of two numbers is 10. Find the numbers if the sum of the their squares is to be a minimum...
I have no idea what exactly the minimum is in this problem...
Thanks!
a minimum of?
or does it mean something like?
1+9 =10
$\displaystyle 1^2+9^2=82$
$\displaystyle 2^2+8^2=68$
I guess you could do it as a trial error like that . . .
or you could do it the "calculus" way:
$\displaystyle x + y = 10$
$\displaystyle y=10-x$
$\displaystyle x^2+y^2$
$\displaystyle x^2+(10-x)^2$
$\displaystyle x^2+x^2-20x+100$
and then take the derivative and see where it goes to zero. *shrugs*
Well first of all let the two numbers be a and b.
Now if their sum is 10, then we can say
$\displaystyle a+b=10$
$\displaystyle b=10-a$
Now the seum of their squares is
$\displaystyle a^2+b^2$
let this equal another variable, say, y
so
$\displaystyle a^2+b^2=y$
substituting in for b from the equation established above, we get
$\displaystyle y=a^2+(10-a)^2$
rearrange
$\displaystyle y=a^2+100-20a+a^2$
$\displaystyle y=2a^2-20a+100$
to find where y has a max/min, we have to fins dy/dx and let it equal 0
$\displaystyle \frac{dy}{dx} = 4a-20$
$\displaystyle 0=4a-20$
$\displaystyle a=5$
There you go, that's one of the two numbers, and the second one can be found with the first formula we established.
This kind of approach is best for these questions, if you're stuck always try and say "let this equal a and that equal b", and see where it takes you.