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