Question

A geometry student wants to draw a rectangle inscribed in the ellipse x²+4y²=36. What is the area of the largest rectangle that the student can draw?

Answer 1

We have an ellipse defined by:

`x^2+4y^2=36`

We have to find what is the area of the largest rectangle inscribed in that ellipse.

As the ellipse is centered in the center of coordinates, we can take the first quadrant definition of the function.

Then, maximize the product x*y. This will show us the maximum area on the first quadrant. As it is the same in all the quadrants, the total area will be that area x*y multiplied by 4, the number of quadrants.

The definition of y=f(x) in the first quadrant can be written as:

`\begin{gathered} x,y>0 \\ x^2+4y^2=36 \\ 4y^2=36-x^2 \\ y^2=-(x^2)/(4)+9 \\ y=\sqrt[]{9-(x^2)/(4)} \end{gathered}`

We have to maximize the value of x*y, as the rectangle will have sides 2x and 2y.

The definiton of g(x)=x*y is:

`g(x)=x\cdot y=x\cdot\sqrt[]{(x^2)/(4)+9}=\sqrt[]{(x^4)/(4)+9x^2}`

To maximize g(x) we can derive it and equal to zero:

`(dg)/(dx)=-(1)/(2)(-(x^4)/(4)+9x^2)^{-(1)/(2)}\cdot(-x^3+18x)=-(1)/(2)\frac{-x^3+18x}{\sqrt[]{(-x^4)/(4)+9x^2}}=0`

The derivative will be 0 if the numerator is 0 (and the denominator is not), so we can write:

`\begin{gathered} -x^3+18x=0 \\ x(-x^2+18)=0 \\ -x^2+18=0 \\ -x^2=-18 \\ x=\sqrt[]{18}=\sqrt[]{9\cdot2}=3\sqrt[]{2} \end{gathered}`

NOTE: x=0 is not a solution, because the denominator is also 0. And x=-3*2^0.5 is not a solution because x must be higher than 0.

We can now calculate the value of y that corresponds to x=3*sqrt(2):

`\begin{gathered} f(3\sqrt[]{2})=\sqrt[]{9-\frac{(3\sqrt[]{2})^2}{4}} \\ f(3\sqrt[]{2})=\sqrt[]{9-(9\cdot2)/(4)} \\ f(3\sqrt[]{2})=\sqrt[]{9-(1)/(2)\cdot9} \\ f(3\sqrt[]{2})=\sqrt[]{(9)/(2)} \\ f(3\sqrt[]{2})=\frac{3}{\sqrt[]{2}} \end{gathered}`

We can now calculate the area x*y as:

`x\cdot y=(3\sqrt[]{2})\cdot\frac{3}{\sqrt[]{2}}=3\cdot3=9`

The area of the inscribed rectangle is 4 times the area x*y, as there are 4 quadrants, so the total area is:

`A=4\cdot xy=4\cdot9=36`

Answer: the maximum area for an inscribed rectangle in the ellipse is 36 units.