Thanks, alan.....here's a non-Calculus qpproach
Let the product xy = a
And xy will be maximized where this function is tangent to the ellipse...so we need to find the "a" that makes this true
So xy = a ⇒ y = a/x
Subbing this into x^2 + 2y^2 = 16 we get
x^2 + 2(a/x)^2 = 16 multiply through by x^2
x^4 + 2a^2 = 16x^2
Let v = x^2
v^2 - 16v + 2a^2 = 0
We will have a double real (positive) vaue for v whenever
256 - 8a^2 = 0
256 = 8a^2
a^2 = 32 ⇒ 2a^2 = 64
a = √32 = 4√2
So
v^2 - 16v - 64 = 0
(v - 8)^2 = 0 ⇒ v = 8 ⇒ x^2 = 8 ⇒ x = ±√8 ⇒ x = ±2√2
And y = (a/x) = (4√2 / ( ±2√2) ) = ±2
So.....the product xy is maximized whenever (x, y) = ( ±2√2, ±2)
And xy is maximized when a = 4√2 as this graph shows :
https://www.desmos.com/calculator/vmykhikdjb