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

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:

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

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

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

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step-by-step explanation: