Question 1190186
(a) The sum of their squares is as large as possible; 
as small as possible
<pre>
Let one number be x and the other be 30-x.
Let y be the sum of the squares of the two numbers.

Since they are non-negative, 

{{{x>=0}}} and {{{30-x>=0}}}
               {{{-x>=-30}}}
               {{{x<=30}}}

Thus the domain of the function is [0,30]

Let y = the sum of their squares  

{{{y=x^2+(30-x)^2}}}

Take the derivatives of both sides:

{{{dy/dx=2x+2(30-x)(-1)}}}

{{{dy/dx=2x-2(30-x)}}}

{{{dy/dx=2x-60+2x}}}

{{{dy/dx=4x-60}}}

Set derivative = 0 to find extremum points

{{{4x-60=0}}}

{{{4x=60}}}

{{{x=15}}}

The candidates for extremum values are this 15 and the endpoints
of the range, 0 and 30

We substitute 0

{{{y=x^2+(30-x)^2}}}
{{{y=0^2+(30-0)^2}}}
{{{y=0+(30)^2}}}
{{{y=900}}}

We substitute 30

{{{y=30^2+(30-30)^2}}}
{{{y=900+(0)^2}}}
{{{y=900+0}}}
{{{y=900}}}

We substitute 15

{{{y=15^2+(30-15)^2}}}
{{{y=225+(15)^2}}}
{{{y=225+225}}}
{{{y=450}}}

So the sum of the squares, which is y, is as large as possible, 900,
when x=0 and when x=30.
The sum of the squares is as small as possible, 450,
when x=15.

----------------------------------------------------

Let y = the square of one number plus the square root of the other
number 

{{{y=x^2+sqrt(30-x)}}}

You can do it the same way, but the going get's really tough.

It has a minimum value at approximately x = 0.0456786468
It has a maximum value at x=30

Edwin</pre>