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:
[tex]x=\sqrt{4-4y^2 }\\[/tex]
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
[tex]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}[/tex]
Therefore the length = 2x = 2√2, the width = 2y = 2/√2
