Question

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

Answer 1

Step 1

Area of rectangle A=2x.2y

x2=16−4y2⇒A2=256y2−64y4

Step 2

A2=(4xy)2=16x2y2

A2=16y2(16−4y2)

256y2−64y4

Step 3

ddy(A2)=512y−256y3−0

⇒y=2–√

x=22–√

Step 4

d2dy2(A2)= -ve ⇒A is maximum.

The sides of rectangle are

42–√ and 22–√