Chapter A5.1, Problem 40E

To determine

To find: the area of the rectangle.

Answer to Problem 40E

The area of the rectangle is $A(\sqrt{2})=4$ .

The length is $2\sqrt{2}$ and the height is $\sqrt{2}$ .

Explanation of Solution

Given:

Ellipse $x^2+4y^2=4$ .

Calculation:

It is known that for any $x$ in the interval [0, 2].

  $y=\sqrt{1-\frac{x^2}{4}}$

The area of the inscribed rectangle (with height 2y and length 2x given by:

  $\begin{array}{l}A(x)=2x\left(2\sqrt{1-\frac{x^2}{4}}\right)\\ =4x\left(\sqrt{1-\frac{x^2}{4}}\right)\end{array}$

Differentiating with respect to $x$

  $A^{\prime }(x)=4\left(\sqrt{1-\frac{x^2}{4}}\right)-\frac{x^2}{\sqrt{1-\frac{x^2}{4}}}$

Needs to maximize the area,

  $\begin{array}{l}\Rightarrow A^{\prime }(x)=0\\ \Rightarrow 4\left(\sqrt{1-\frac{x^2}{4}}\right)-\frac{x^2}{\sqrt{1-\frac{x^2}{4}}}=0\\ \Rightarrow x=\sqrt{2}\end{array}$

Therefore, $\text{A}(0)=\text{A}(2)=0,$ thus $A(\sqrt{2})=4$ is the maximum area, when the length is $2\sqrt{2}$ and the height is $\sqrt{2}$ .

Conclusion:

Therefore, the area of the rectangle is $A(\sqrt{2})=4$ .

The length is $2\sqrt{2}$ and the height is $\sqrt{2}$ .