SOLUTION: The sum of two non-negative numbers is 30. Find the numbers if: (a) The sum of their squares is as large as possible; as small as possible (b) The square of one number plus

Let one number be 15+x;  then the other number is 15-x.



The sum of their squares is then

    (15+x)^2 + (15-x)^2 = (225 + 2x + x^2) + (225 - 2x + x^2) = 450 + 2x^2.



The addend 2x^2 is always non-negative and is minimal at x= 0.

The sum of the squares is as small as possible at x= 0, when the sum of the squares is 450.

This minimum is provided when both the numbers are equal to 15.



The maximum of the sum of squares is reached at x = +/-15, when x^2 is as large as possible.

Under this condition, the numbers itself are 0 and 30.

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, 

 and 
               
               

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

Let y = the sum of their squares  



Take the derivatives of both sides:









Set derivative = 0 to find extremum points







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

We substitute 0






We substitute 30






We substitute 15






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.

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



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