Question 1088908
{{{x^2 + 2y^2 = 16}}} represents an ellipse centered at (0,0),
which is symmetrical with respect the the x- and y-axes,
and symmetrical with respect to the origin:
{{{drawing(300,300,-4.4,4.4,-4.4,4.4,
grid(0),arc(0,0,8,4sqrt(2),0,360),
circle(2.83,2,0.1),circle(-2.83,-2,0.1),
locate(3,2.3,M),locate(-3.2,-2,P) )}}} .
There will be a maximum for {{{x*y}}} in the first quadrant, at {{{M(x[M],y[M])}}} ,
and another one at {{{P(-x[M],-y[M])}}} , in quadrant III.
All we have to do is find two positive numbers, the coordinates of {{{M}}} .
{{{x^2 + 2y^2 = 16}}}
{{{x^2 = 16 - 2y^2}}}
{{{x^2 y^2= (16 - 2y^2)y^2}}}
{{{(xy)^2= 16y^2 - 2y^4}}}
We are looking for a value {{{y=y[M]>0}}} that makes {{{16y^2 - 2y^4}}} maximum.
If we say {{{t=y^2}}}, the expression becomes {{{16t-2t^2}}} ,
a quadratic polynomial/function:
{{{graph(300,300,-1,9,-10,40,16x-2x^2)}}} .
If you like applying formulas (or if your teacher likes to see formulas),
you may have been told in class that
{{{f(x)=ax^2+bx+c}}} is maximum when {{{x=(-b)/"2 a"}}}
In this case, it means that {{{16t-2t^2}}} is maximum
when {{{t=(-16)/(2(-2))=(-16)/(-4)=4}}} .
 
Otherwise, you may realize that
{{{16t-2t^2=-2(t^2-8t+16-16)=-2(t^2-8t+16)+32=-2(t-4)^2+32<=32}}} .
That tells you that the maximum of {{{(xy)^2}}}is {{{32}}} ,
meaning that the maximum of {{{xy}}} is {{{sqrt(32)=highlight(4sqrt(2))}}} .
 
If you continue working with formulas, {{{t=4}}}  gives you {{{y[M]^2=4}}} --> {{{y[M]=2}}} ,
and {{{x[M]^2 = 16 - 2y[M]^2=16-2*4=16-8=8}}} --> {{{x[M]=sqrt(8)=2sqrt(2)}}} .
Multiplying, you get
{{{x[M]y[M]=(2sqrt(2))*2=highlight(4sqrt(2))=(-x[M])(-y[M])}}} .
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