Maximum area of a rectangle inscribed under y=1-x^2 and above the x-axis.
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In this optimization problem, we
In this optimization problem, we use the derivative to find the maximum area of a rectangle inscribed under y=1-x^2 and above the x-axis. We begin with an animation to illustrate why we are guaranteed to have a maximum value of area for some x between 0 and 1, and we use the computer animation to approximate the location of the maximum value of area and the maximum area of the rectangle at that value of x.
Next, we label the diagram of the inscribed rectangle with a height of 1-x^2 and a width of 2x. This means the area of the rectangle as a function of x is given by A(x)=2x(1-x^2). Taking the derivative and setting equal to zero, we find that the maximum value of area occurs at x=1/sqrt(3), and evaluating A(1/sqrt(3)), we find the maximum area of a rectangle under the parabola y=1-x^2 as 4sqrt(3)/9. Both of these exact values match well to the computer approximations shown at the beginning of the video.
Next, we label the diagram of the inscribed rectangle with a height of 1-x^2 and a width of 2x. This means the area of the rectangle as a function of x is given by A(x)=2x(1-x^2). Taking the derivative and setting equal to zero, we find that the maximum value of area occurs at x=1/sqrt(3), and evaluating A(1/sqrt(3)), we find the maximum area of a rectangle under the parabola y=1-x^2 as 4sqrt(3)/9. Both of these exact values match well to the computer approximations shown at the beginning of the video.
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در تاریخ 1403/02/08 منتشر شده
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