Question

Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse 4x^2 16y^2

Answer 1

Answer:

The answer is below

Step-by-step explanation:

Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse 4x² + 16y² = 16

Solution:

Given that the ellipse has the equation: 4x² + 16y² = 16

let us make x the subject of the formula, hence:

4x² + 16y² = 16

4x² = 16 - 16y²

Dividing through by 4:

x² = (16 - 16y²)/4

x² = 4 - 4y²

Taking square root of both sides:

$x=\sqrt{4-4y^2 }$

The points of the rectangle vertices is at (x,y), (-x,y), (x,-y), (-x,-y). Hence the rectangle has length and width of 2x and 2y.

The area of a rectangle inscribed inside an ellipse is given by:

Area (A) = 4xy

A = 4xy

$A=4(\sqrt{4-4y^2} )y\\\\A=4y\sqrt{4-4y^2}=4\sqrt{4y^2-4y^4} \\\\The\ maximum\ area\ of\ the\ rectangle\ is\ at\ \frac{dA}{dy}=0\\\\ \frac{dA}{dy}=4(\frac{4-8y^2}{\sqrt{4-4y^2} } )\\\\4(\frac{4-8y^2}{\sqrt{4-4y^2} } )=0\\\\4-8y^2=0\\\\8y^2=4\\\\y^2=1/2\\\\y=\frac{1}{\sqrt{2} }\\\\x=\sqrt{4-4(\frac{1}{\sqrt{2} })^2}=\sqrt{2}$

Therefore the length = 2x = 2√2, the width = 2y = 2/√2