m+n=30
m^2+n^2 = m^2+(30-m)^2 = 2m^2 - 60m + 900
Now, that's a parabola with vertex at (15,450)
Since the parabola opens upward, 450 is the minimum, and any maximum will be at the ends of the interval. We can't have m=0, so pick the numbers 1 and 29.
1^2+29^2 = 842
Two non-negative numbers are chosen such that their sum is 30. Find the numbers if the sum of their squares is to be a maximum.
Explanation to solve this would help !
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