Consider the ellipse Find the maximum value of the product on the ellipse.

x^2 + 2y^2 = 16   solve for y..... subtract x^2 from both sides

2y^2 = 16 - x^2     divide both sides by 2

y^2 = 8 -x^2/2      take the positive root of both sides

y = (8 - x^2/2)^(1/2)

Now  we wish to maximize xy....so we have

x(8 - x^2/2)^(1/2)    take the derivative and set this to 0

(8 - x^2/2)^(1/2) -(1/2)(x^2)(8 - x^2/2)^(-1/2) = 0   factor

(8- x^2)^(-1/2)[(8 - x^2/2) -(1/2)(x^2)] = 0

So we have

[(8 - x^2/2) -(1/2)(x^2)] = 0

8 - x^2 = 0

x^2 = 8

x =√8  = 2√2

So y = (8 - 8/2)^(1/2) = (8 - 4)^(1/2) = (4)^(1/2) = 2

And the max product =   (2√2)(2) = 4√2 = about 5.65685

Nice one Chris!

Here is the graph.

https://www.desmos.com/calculator/t35djkjcd8

Thanks, Melody.......