Can someone please help me set up this problem?
Find the larger of two numbers whose sum is 30 for which the sum of their squares is a minimum.
Let the first number $\displaystyle x_1$ and the second number $\displaystyle x_2$.
Then $\displaystyle f\left( {x_1 ,x_2 } \right) = x_1^2 + x_2^2 \to \min$ and $\displaystyle x_1+x_2=30\Rightarrow$.
$\displaystyle \Rightarrow x_2=30-x_1\Rightarrow f\left( {x_1 } \right) = x_1^2 + \left( {30-x_1} \right)^2 \to \min$
Do you undestand?
Denote these two numbers to be $\displaystyle x$ and $\displaystyle y$.
#1: Their sum must be 30, so: $\displaystyle x+y=30$, or $\displaystyle y=30-x$
#2: And you want to minimize: $\displaystyle x^2+y^2$
Given #1 as a constraint, #2 becomes: $\displaystyle x^2+(30-x)^2$. Differentiate this expression, set it equal to zero (to minimize), and find $\displaystyle x$