A rectangle is bounded by the x-axis and the semicircle y= �ã(25-x^2). What length and width should the rectangle have so that the area is maximum?
Place the semi-circle so it's center coincides with the origin. If you go a distance x to the right the max height is sqrt(25-x^2). Since x is only half the length of one side, the dimensions are 2x and sqrt(25-x^2), thus the area is
A=2x*sqrt(25-x^2)
Find dA/dx, set to 0 and solve for x.
To find the length and width of the rectangle that will maximize its area, we need to find the first derivative of the area equation and set it equal to zero. Let's go through the steps to solve this problem:
1. Place the semicircle so that its center coincides with the origin. This means the equation of the semicircle is y = √(25 - x^2).
2. Consider that x represents half the length of one side of the rectangle, and the height of the rectangle is the same as the y-value of the semicircle.
3. Therefore, the length of the rectangle is 2x (twice the value of x), and the width is √(25 - x^2).
4. The area of the rectangle is given by the equation A = 2x * √(25 - x^2).
5. Now, let's find the derivative of the area function with respect to x. We use the product rule here:
dA/dx = 2 * (√(25 - x^2)) + 2x * (1/2) * (25 - x^2)^(-1/2) * (-2x)
Simplifying the derivative, we get:
dA/dx = 2 * √(25 - x^2) - 2x^2 / √(25 - x^2)
6. Set dA/dx equal to zero and solve for x:
2 * √(25 - x^2) - 2x^2 / √(25 - x^2) = 0
Multiplying both sides by √(25 - x^2), we get:
2 * (25 - x^2) - 2x^2 = 0
Simplifying further:
50 - 2x^2 - 2x^2 = 0
50 - 4x^2 = 0
4x^2 = 50
x^2 = 50 / 4
x^2 = 12.5
x = ±√(12.5)
Since the width cannot be negative, we take the positive square root:
x = √(12.5)
7. Now, substitute the value of x back into the equation for the width to find its value:
Width = √(25 - (√(12.5))^2)
Width = √(25 - 12.5)
Width = √12.5
8. The length of the rectangle is then twice the value of x:
Length = 2 * √(12.5)
Length = 2√12.5
9. Therefore, to maximize the area of the rectangle, it should have a length of 2√12.5 and a width of √12.5.
Remember, these calculations assume that the semicircle's center coincides with the origin. If the semicircle is shifted, you would need to adjust the calculations accordingly.