+1207 Consider all the points in the plane that solve the equation \(x^2 + 2y^2 = 16.\) Find the maximum value of the product xy on this graph.
+129852 x^2 + 2y^2 = 16
2y^2 = 16 - x^2
y^2 = (16 - x^2)/2 take the positive root on both sides
y = [ (1/2)(16 - x^2)]^(1/2) = [ 8 - x^2/2]^(1/2)
xy = x [ 8 - x^2/2]^(1/2) take the derivative of this with respect to x and set to 0
y ' = [ 8 - x^2/2]^(1/2) - (x)(1/2) [ 8 - x^2/2] ^(-1/2) (x) =
[8 - x^2/2]^(1/2) - (x^2/2) [ 8 - x^2/2]^(-1/2) = 0 factor
[8 -x^2/2]^(-1/2) [ [8-x^2/2 -x^2/2] = 0
[8 -x^2/2]^(-1/2) [ 8 - x^2] = 0 divide both sides by the first factor and we have that
8 - x^2 = 0
x^2 = 8 take the positive root
x = √8 = 2√2
And y = √ [ 8 - (√8)^2 / 2 ] = √[ 8 -8/2] = √ 8 -4] = √4 = 2
So.....the max value of xy on the graph is (2√2) (2) = 4√2
