Find the largest area of the largest rectangle that can be cut from a circular quadrant.

Question

A rectangle with sides parallel to the coordinate axes is inscribed in

the ellipse.

x^2/a^2+y^2/b^2=1

Find the largest possible area for this rectangle.

been stuck on this problem for a while and can't get a grasp on it.

Answer 1

Any point P on the ellipse can be represented as (a·cos(θ), b·sin(θ)), where θ can range from 0 to 2π.

Let us assume A is the corner of the inscribed rectangle that is located in the first quadrant. Then θ only ranges from 0 to π ⁄ 2.

Then, the lengths of the sides of the inscribed rectangle are 2a·cos(θ) and 2b·sin(θ). And the area of the rectangle is 4ab·cos(θ)·sin(θ) = 2ab·sin(2θ) since 2·sin(θ)·cos(θ) = sin(2θ).

In summary, the area of the rectangle is 2ab·sin(2θ) which in turn has a maximum value of 2ab since sin(2θ) has a maximum value of 1.

Answer : 2ab.

Hope that helps!

Answer 2

If you draw a picture of the inscribed rectangle, the area is made up of 4 squares. The dimensions of each

square is x by y. We only need to maximize one square and multiply its area by 4.

Area of one square is:

A = xy

y = ab√(a²-x²)

A(x) = abx√(a²-x²)

To maximize, find dA/dx, set that to zero and solve for x.

A'(x) = abx(-x/√(a²-x²))+ab√(a²-x²)

Setting to zero and simplifying:

x²/√(a²-x²) = √(a²-x²)

x²=a²-x²

x = a/√2

and y = b√2

So the maximum area of the entire inscribed rectangle is:

4xy = 4(a/√2)(b/√2) = 2ab

You can use the 2nd derivative test to show this is a maximum.