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

Toolbox:
• For greatest area, maximise area.
• Working rule for maxima minima.
• Find the function to be maximised or minimised
• Convert the function into single variable
• Apply the conditions for maximum or minimum i.e., first derivative = 0 ( w.r.to the variable converted) and second derivative positive for minima and negative for maxima.
Step 1
Area of rectangle $A = 2x.2y$
$x^2=16-4y^2 \Rightarrow A^2=256y^2-64y^4$
Step 2
$A^2 = (4xy)^2 = 16x^2y^2$
$A^2 = 16y^2(16-4y^2)$
$256y^2-64y^4$
Step 3
$\large\frac{d}{dy}(A^2)=512y-256y^3-0$
$\Rightarrow y = \sqrt 2$
$x = 2\sqrt 2$
Step 4
$\large\frac{d^2}{dy^2}(A^2)=$ -ve $\Rightarrow A$ is maximum.
The sides of rectangle are
$4\sqrt 2$ and $2\sqrt 2$