document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #848212 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "I'll refer to the two nonnegative numbers as m and n.\r
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\n" ); document.write( "\n" ); document.write( "m^2 + 5n^2 = 30 rearranges to m^2 = 30 - 5n^2\r
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\n" ); document.write( "\n" ); document.write( "Plug that into the expression we want to max out.
\n" ); document.write( "m^2 + n^2
\n" ); document.write( "= 30-5n^2 + n^2
\n" ); document.write( "= 30-4n^2\r
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\n" ); document.write( "\n" ); document.write( "We want to make 30-4n^2 as large as possible.
\n" ); document.write( "This is equivalent to looking for the highest point on the upside down parabola y = 30-4x^2 aka y = -4x^2+30.\r
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\n" ); document.write( "\n" ); document.write( "The smallest that x^2 can get is 0.
\n" ); document.write( "The negative out front will flip things to show that -4x^2 maxes out when x = 0
\n" ); document.write( "Therefore y = -4x^2+30 maxes out when x = 0 and its paired value is y = 30.\r
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\n" ); document.write( "\n" ); document.write( "The vertex of y = -4x^2+30 is (0,30) which is the highest point on this parabola.
\n" ); document.write( "You can use a graphing tool such as Desmos and GeoGebra to confirm.
\n" ); document.write( "Or you can go old-school with something like a TI83.\r
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\n" ); document.write( "\n" ); document.write( "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.
\n" ); document.write( "A somewhat similar question is found here\r
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\n" ); document.write( "\n" ); document.write( "Answer: 30
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