**Problem:**

Find the rectangle of largest area inscribed in the ellispse:

`> `
**ellipse := (x/3)^2 + (y/2)^2=1;**

**Solution:**

By the obvious symmetry of the ellipse we only need to consider rectangles

with sides parallel to the axes and look only at the first quadrant. Thus, we

need to find the 0 < x < 3 that produces the rectangle with largest area as illustrated

by the following picture,

`> `
**with(plots):**

Warning, the name changecoords has been redefined

`> `
**p1 := plot(2*sqrt(1-x^2/9),x=0...4,y=0..4):**

`> `
**p2 := plot(2*sqrt(1-4/9),x=0..2,color=black):**

`> `
**p3 := plot([2,t,t=0..2*sqrt(1-4/9)],color=black,xtickmarks=[0,3]):**

`> `
**display({p1,p2,p3});**

For a given value of x the area is,

`> `
**A := x*h(x):**

with h(x) obtained from the positive solution of y in terms of x,

`> `
**h := solve(ellipse,y);**

`> `
**A := x*h[1];**

The Area function clearly has a single maximum around x about 2 as it shown by the picture,

`> `
**plot(A,x=0..3);**

To obtain the exact maximum, we just find the place where the derivative of A is 0.

`> `
**xsol := solve(diff(A,x));**

and only the positive solution makes sense here so,

`> `
**xsol := xsol[2]; evalf(xsol);**

The optimal rectangle has total area,

`> `
**Aopt := simplify(4*subs(x=xsol,A));**

for comparison, the area of the ellipse is,

`> `
**Aell := Pi*3*2; evalf(Aell);**

Hence, the rectangle with largest area covers,

`> `
**PercentCovered := evalf(Aopt/Aell)*100;**

`> `

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