Any help would be appreciated!

Question

A rectangle has one side on the 𝑥x-axis and two vertices on the curve

y=7/(1+x^2)

Find the vertices of the rectangle with maximum area. 

Answer 1

Area A(x) = 2x·7/(1 + ,x²) = 14x/(1 + x²).

A'(x) = 14(1 + x² - 2x²)/(1 + x²)² + 14(1 - x²)/(1 + x²)² = 0; 1 - x² = 0; x = ± 1

At x = 1 A(x) has max because derivative from left to right chenges sign from + to - .

So verteces of maximum area rectangle: (-1, 0), (-1, 7/2), (1, 7/2), (1, 0)

Answer 2

Here's a graph showing the rectangle described:

https://www.desmos.com/calculator/q7spcj3pef

It's area can be calculated using AREA - A(x) = 2x(7/(1+x^2) = 14x/ (1 +x^2)

In order to maximize the AREA function, we will differentiate it and calculate when A'(x) = 0

A'(x) = {14(1 +x^2) - 14x(2x) ] / (1+x^2)^2

14 + 14x^2 - 28x^2 = 0

14 - 42x^2 = 0

x^2 = 14/42 = 1/3

the x value that gives the maximum area is the square root of 1/3

https://www.desmos.com/calculator/stucsdvwsw