Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse (in Description below) with sides parallel to the coordinate axes.

Question

Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse (x²/25)+(y²/9)=1 with sides parallel to the coordinate axes.

Let length be the dimension parallel to the​ x-axis and let width be the dimension parallel to the​ y-axis.

Answer 1

Let horizontal side is 2x, then vertical side is 2y. So area is f(x,y) = 4xy and under condition

x²/25 + y²/9 = 1;

Lagrange function: F(x, y) = 4xy - λ(x²/25 + y²/9 - 1)

F_x = 4y - 2λx/25 = 0; λ = 50y/x;

F_y = 4x - 2λy/9 = 0; λ = 18x/y; 50y/x = 18x/y; 25y² = 9x²; y² = 9x²/25

F_λ = - x²/25 - y²/9 + 1 = 0; x²/25 + x²/25 = 1; x² = 25/2; x = ±5√2/2

y² = 9·(25/2)/25 = 9/2; y = ±3√2/2. so, Area is A = 4·5√2/2·3√2/2 = 30.

To show that this is max area we take different point on ellipse as vertex of rectangle for example x = 3

then y = 12/5 and area a = 4·3·12/5 = 144/5 < 30