Question 1209307
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I'll refer to the two nonnegative numbers as m and n.


m^2 + 5n^2 = 30 rearranges to m^2 = 30 - 5n^2


Plug that into the expression we want to max out.
m^2 + n^2
= 30-5n^2 + n^2
= 30-4n^2


We want to make 30-4n^2 as large as possible.
This is equivalent to looking for the highest point on the upside down parabola y = 30-4x^2 aka y = -4x^2+30.


The smallest that x^2 can get is 0.
The negative out front will flip things to show that -4x^2 maxes out when x = 0
Therefore y = -4x^2+30 maxes out when x = 0 and its paired value is y = 30.


The vertex of y = -4x^2+30 is (0,30) which is the highest point on this parabola.
You can use a graphing tool such as <a href="https://www.desmos.com/calculator">Desmos</a> and <a href="https://www.geogebra.org/calculator">GeoGebra</a> to confirm. 
Or you can go old-school with something like a TI83.


The y coordinate of this vertex is the max value of 30-4n^2 aka m^2 + n^2 when m^2+5n^2 = 30.
A somewhat similar question is found <a href="https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Quadratic-relations-and-conic-sections.faq.question.1208800.html">here</a>



Answer: <font color=red>30</font>
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