Question

) a rectangle has one corner at the point (0, 0), a second corner on the x-axis, a third corner on the y-axis, and the fourth corner lying on the line segment between (0, 1) and (2, 0), as shown below. what is the maximum area of the rectangle?

Answer 1

The maximum value of x^2/2 occurs at x=2, and the maximum area of the rectangle is 2^2/2​ = 2 .

Let's denote the point where the rectangle intersects the line segment as (x,y).

Then, the area of the rectangle is xy, and we want to maximize this.

The equation of the line segment is y=1− x/2 , so we can substitute this into xy to get xy=x(1− x/2 )= x^2/2 .

Then, we want to maximize the quantity x^2/2 subject to the constraint that 0≤x≤2.

The derivative of x^2/2 is x, which is positive for x>0 and negative for x<0. This means that x^2/2 is increasing on the interval [0,2].

Therefore, the maximum value of x^2/2 occurs at x=2, and the maximum area of the rectangle is 2^2/2​ = 2 .