Question 1208800
<font color=black size=3>
Answer: <font color=red>6</font>



Explanation


Let's solve for y in the given equation
x^2 + 5y^2 = 30
5y^2 = 30 - x^2
y^2 = (30 - x^2)/5
y^2 = 6 - 0.2x^2
y = sqrt(6 - 0.2x^2)
Since y is nonnegative, we ignore the negative branch.



The expression
x+y
turns into
x + sqrt(6 - 0.2x^2)


If you were to graph this, and use the calculator's "maximum" feature, then you should find the highest point is located at (5,<font color=red>6</font>). 
This shows that the largest value of x+y is <font color=red>6</font>
{{{
drawing(500,500,-7,7,-7,7,
graph(500,500,-7,7,-7,7,-100,x + sqrt(6 - 0.2x^2)),
circle(5,6,0.08),circle(5,6,0.1),circle(5,6,0.12),circle(5,6,0.14),
locate(5-0.3,6+0.6,"(5,6)")
)
}}}
Graph of f(x) = x + sqrt(6 - 0.2x^2)
<a href="https://www.geogebra.org/calculator">GeoGebra</a> and <a href="https://www.desmos.com/calculator">Desmos</a> are two of many graphing tools to use. 
Another way to determine this highest point is to use differential calculus. 


Caution: Do not mix up the original y value with the output of the function x + sqrt(6 - 0.2x^2)


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Another approach.


Consider these two sets of numbers
{ a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..., a<sub>n</sub> }
{ b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub>, ..., b<sub>n</sub> }


Due to the <a href="https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality">Cauchy-Schwarz Inequality</a> we can state the following
{{{(matrix(1,3,a[1]^2+a[2]^2,"+...+",a[n]^2))(matrix(1,3,b[1]^2+b[2]^2,"+...+",b[n]^2)) >= (matrix(1,3,a[1]b[1]+a[2]b[2],"+...+",a[n]b[n]))^2}}}


For new students, and perhaps some experienced students as well, this inequality may seem intimidating. 


Luckily we only need to worry about sets with 2 items each
{ a<sub>1</sub>, a<sub>2</sub> }
{ b<sub>1</sub>, b<sub>2</sub> }


Allowing us to use an inequality that's far more easier to work with.
{{{(a[1]^2+a[2]^2)(b[1]^2+b[2]^2) >= (a[1]b[1]+a[2]b[2])^2}}}


Let
{{{a[1] = x}}} and {{{a[2] = y*sqrt(5)}}}
Square both sides of each equation.
{{{a[1]^2 = x^2}}} and {{{a[2]^2 = 5y^2}}}
The selection of a<sub>1</sub> & a<sub>2</sub> is to set up x^2+5y^2.


So,
{{{(a[1]^2+a[2]^2)(b[1]^2+b[2]^2) >= (a[1]b[1]+a[2]b[2])^2}}}


{{{(x^2+5y^2)(b[1]^2+b[2]^2) >= (x*b[1]+y*sqrt(5)*b[2])^2}}}


Now we want x+y to show up on the right hand side.
To do this we can say {{{b[1] = 1}}} and {{{b[2] = 1/sqrt(5)}}}


{{{(x^2+5y^2)(b[1]^2+b[2]^2) >= (x*b[1]+y*sqrt(5)*b[2])^2}}}


{{{(x^2+5y^2)(1^2+(1/sqrt(5))^2) >= (x*1+y*sqrt(5)*(1/sqrt(5)))^2}}}


{{{(x^2+5y^2)(1 + 1/5) >= (x+y)^2}}}


{{{(x^2+5y^2)(6/5) >= (x+y)^2}}}


{{{(30)(6/5) >= (x+y)^2}}} ...... use the fact that x^2+5y^2 = 30 is given to us


{{{36 >= (x+y)^2}}}


{{{(x+y)^2 <= 36}}}


{{{x+y <= sqrt(36)}}}


{{{x+y <= 6}}}


This shows that x+y <font color=red>maxes out at 6</font>


Note that because x,y are nonnegative, we dont need to worry about the plus minus. We focus entirely on the positive square root.


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Yet another approach.


This approach is inspired by tutor ikleyn.
Graph x^2 + 5y^2 = 30 which is an ellipse and graph x+y = c, where c is a parameter.


x+y = c is a family of lines such as x+y=1, x+y=1.5, x+y=1.8, etc.


Here is an interactive Desmos graph where you can move the slider around for c.
<a href="https://www.desmos.com/calculator/zyi3ize89w">https://www.desmos.com/calculator/zyi3ize89w</a>
<font color=red>c = 6</font> is the largest possible value if we want the blue line to intersect with the red ellipse. The point of tangency is (x,y) = (5,1)
When c > 6, the blue line no longer intersects the ellipse.
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