Question 175552
The first part of this problem is interpretation.

two numbers whose sum is 30 means {{{x + y = 30}}} , since we don't know wither number we need two variables.

the square of one number plus ten times the other number means {{{f(x)=x^2 + 10y}}} I am calling the equation f(x) so we have a name for it.

The next thing to remember is that to find the minimum of a function you take the derivative(calculus), set the derivative equal and solve. 

Since they wanted a minimum for "such that the sum of the square of one number plus ten times the other number is a minimum" we need to take the derivative of f(x)

This presents a small problem since you can only have one variable when taking the derivative. We solve this by rewriting {{{x + y = 30}}} as {{{y=30-x}}}

We then substitute this into f(x), giving us {{{ f(x)=x^2 + 10(30-x)}}} I replaced the {{{y}}} with {{{ 30-x}}}. 
We now have {{{f(x)= x^2 +300-10x}}} 

The next step is to take to derivative. {{{ (d/dx) f(x)=2x-10 }}}, the derivative of {{{x^2}}} was {{{2x}}}, the derivative of 300 since t is just a number is 0, and the derivative of {{{-10x}}} is -10, giving you {{{2x-10}}} as your derivative

You then take {{{2x-10=0}}} and solve
{{{2x=10}}}
and
{{{x=10/2=5}}}

Plugging back into {{{x+y=30}}} you now have {{{5+y=30}}} giving you {{{y=25}}}

So the answer is x=5 and y=25 will give you the minimum