SOLUTION: Let x and y be nonnegative real numbers. If x^2 + 5y^2 = 30, then find the maximum value of x^2 + y^2.

Question 1209307: Let x and y be nonnegative real numbers. If x^2 + 5y^2 = 30, then find the maximum value of x^2 + y^2.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
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Let x and y be nonnegative real numbers.
If x^2 + 5y^2 = 30, then find the maximum value of x^2 + y^2.
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From the given equation x^2 + 5y^2 = 30, express x^2 = 30 - 5y^2 and substitute it into another expression. You will get x^2 + y^2 = (30-5y^2) + y^2 = 30 - 4y^2. It has the maximum value of 30 when y = 0 (which is a non-negative real number). ANSWER. The requested maximum value of x^2+y^2 is 30.

Solved.

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

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 Desmos and GeoGebra 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 here

Answer: 30

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