Question

A rectangle has one side on the xx-axis and two vertices on the curve y=74+x2. y=74+x2. find the vertices of the rectangle with maximum area.

Answer 1

Given that a rectangle has one side on the x-axis and two vertices on the curve $y= \frac{7}{4+x^2}$

We find the vertices of the rectangle with maximum area as follows:

Because of the symmetry of the given curve, the base of the retangle is 2x and the height of the rectangle is y.

Thus, the area of the required rectangle is 2xy where $y= \frac{7}{4+x^2}$

Thus, $Area=\frac{14x}{4+x^2}$

For maximum area, we find value of x for which the derivative of the function for the area is equal to zero.

The derivative of the function for the area is given by:

$(Area)'= \frac{14(4-x^2)}{(4+x^2)^2} =0 \\ \\ \Rightarrow4-x^2=0 \\ \\ \Rightarrow x^2=4 \\ \\ \Rightarrow x=\sqrt{4}=2$

Thus, the area is maximum when x = -2 and x = 2

When x = -2,

$y= \frac{7}{4+(-2)^2}= \frac{7}{4+4} = \frac{7}{8}$

Similarly, when x = 2,

$y= \frac{7}{4+(2)^2}= \frac{7}{4+4} = \frac{7}{8}$

Therefore, the required vertices are: $(-2,0),\ \left(-2, \frac{7}{8} \right), \ \left(2, \frac{7}{8}\right), \ (2, 0)$