Question

Find the sides of a rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=16$

Answer 1

Let the one corner of rectangle be at $(x,y)$

Then the other corners will be at $(x,-y),(-x,y),(-x,-y)$

Sides of the rectangle are $2x,2y$

Then the area of the rectangle $A=2x\times 2y=4xy$

$⟹A^2=16x^2{y}^2$

Given $(x,y)$ lies on the ellipse $x^2+4y^2=16⟹x^2=16-4y^2$

So $A^2=16(16-4y^2)y^2=256y^2-64y^4$

Differentiating on both sides

$2A\frac{dA}{dx}=512y-256y^3=0⟹y^2=2⟹y=\pm \sqrt{2}$

$x^2=16-4y^2=16-8=8⟹x=\pm 2\sqrt{2}$

so the sides are $4\sqrt{2},2\sqrt{2}$