Maximum area using calculus?

Question

Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes, that can be inscribed in the ellipse given by:

(x^2)/144 + (y^2)/16 = 1

Answer 1

area of an inscribed rectangle is 4xy where (x,y) is a point on the ellipse.

y = 4*sqrt(1 - (x/12)^2)

Area = 16 x sqrt(1 - (x/12)^2)
to find maximum area set the derivative of A wrt x to 0

dA/dx = -8x (2x/144) /sqrt(1 - (x/12)^2) + 16*sqrt(1 - (x/12)^2)
setting the above to 0 and solving for x gives

(x/12)^2 = 1 - (x/12)^2
x = 6sqrt(2)

y = 4*sqrt(1 - (1/2)) = 2sqrt(2)

Maximum Area = 4xy = 24*4 = 96

Answer 2

Let A(x,y) = (2x)(2y) = 4 xy be the area of the rectangle (assuming the rectangle will be centered at the origin).

We want to maxize A subject to the the constraint

g(x,y) = x^2/144 + y^2/16 = 1.

A will be maximal when f(x,y) = xy is maximal, so to keep the numbers small let's maximize f first.

Using the Lagrange multipliers method, we look for where the gradient of f is parallel to the gradient of g. i.e. we need to solve the system

f_x(x,y) = c g_x(x,y); f_y(x,y) = c g_y(x,y); g(x,y) = 1.

for c, x, and y (really just x and y; unless finding c helps us find x and y).

Taking the partial derivatives, the first two equations become

y = c 2x/144, x = c 2y/16,
72y/x = c, 8x/y = c
72y/x = 8x/y
72 y^2 = 8 x^2
9 y^2 = x^2
3 y = x (we can assume x and y are positive)

Now we use this in the constraint:
x^2/144 + y^2/16 = 1
9y^2/144 + y^2/16 = 1
y^2 + y^2 = 16
y^2 = 8
y = sqrt(8)
x = 3 sqrt(8)

So the maximal rectangle will be 2sqrt(8) by 6 sqrt(8), with an area of 2sqrt(8)*6sqrt(8) = 96

Answer 3

This is the equation of an ellipse centered on the origin.

x²/144 + y²/16 = 1

Multiply by 144 to clear the denominators.

x² + 9y² = 144
x² = 144 - 9y²
x = √(144 - 9y²)
____________

Let
A = area rectangle

The dimensions of the rectangle are 2x by 2y.

A = (2x)(2y) = 4xy
A = 4y√(144 - 9y²)

Take the derivative and set it equal to zero to find the critical point(s).

dA/dy = 4√(144 - 9y²) + (1/2)(-9*2y)(4y)/√(144 - 9y²) = 0
4√(144 - 9y²) - 36y²/√(144 - 9y²) = 0
√(144 - 9y²) - 9y²/√(144 - 9y²) = 0
√(144 - 9y²) = 9y²/√(144 - 9y²)

144 - 9y² = 9y²
144 = 18y²
8 = y²
y = √8 = 2√2

Solve for x.

x = √(144 - 9y²)
x = √(144 - 9*8) = √(144 - 72) = √72
x = 6√2

The dimensions of the rectangle are

2x by 2y = 12√2 by 4√2