A rectangle is bounded by the x-axis and the semicircle y=�ã(25-x^2).
Question is, what length and width should the rectangle have so that its area is a maximum, and what is the maxuimum area?
Area= length*width
= 2x*y= 2x*sqrt(25-x^2)
Now, take that, differentiate it, set to zero, and solve for x,y. Length = 2x, width (or height) is y.
I will be happy to critique your work or thinking.
Differentiating with respect to x:
2*sqrt(25-x^2) - 4x*(1/2)*(25-x^2)^(-1/2) = 0
2*sqrt(25-x^2) = 4x*(1/2)*(25-x^2)^(-1/2)
4x^2 = 25-x^2
5x^2 = 25
x = sqrt(25/5)
y = sqrt(25-x^2)
= sqrt(25-(25/5))
= sqrt(20/5)
Therefore, the length and width of the rectangle should be 2*sqrt(25/5) and sqrt(20/5) respectively, and the maximum area is (2*sqrt(25/5))*(sqrt(20/5)) = 4*sqrt(500/25) = 4*sqrt(20) units^2.
To find the dimensions that maximize the area of the rectangle, we need to maximize the function representing the area. In this case, the area function is given by:
Area = 2x * √(25 - x^2)
To find the maximum area, we can differentiate the area function with respect to x and set it equal to zero.
d(Area)/dx = 2 * √(25 - x^2) + 2x * (1/2) * (25 - x^2)^(-1/2) (-2x)
Simplifying this expression, we get:
2 * √(25 - x^2) - (2x^2) / √(25 - x^2) = 0
Multiplying through by √(25 - x^2), we get:
2(25 - x^2) - 2x^2 = 0
Expanding and rearranging, we have:
50 - 2x^2 - 2x^2 = 0
4x^2 = 50
x^2 = 50/4
x^2 = 12.5
Taking the positive square root (since dimensions cannot be negative), we have:
x = √12.5
Now, we can substitute this value of x back into the area function to find the corresponding value of y:
Area = 2x * √(25 - x^2)
Substituting x = √12.5, we have:
Area = 2 * √12.5 * √(25 - 12.5)
Area = 2 * √(12.5) * √(12.5)
Area = 2 * 3.5355 * 3.5355
Area ≈ 24.999
Therefore, the maximum area of the rectangle is approximately 24.999 square units. The length and width of the rectangle are approximately equal to 2√12.5 units.
To find the dimensions of the rectangle that would maximize its area, we need to differentiate the equation for the area with respect to x and set it equal to zero.
Let's start by differentiating the expression for the area:
A = 2x * sqrt(25 - x^2)
To simplify the differentiation process, let's rewrite the expression as:
A = 2x * (25 - x^2)^(1/2)
Now, using the product rule of differentiation, we can find the derivative of A with respect to x:
dA/dx = 2 * sqrt(25 - x^2) - 2x * (1/2) * (25 - x^2)^(-1/2) * (-2x)
Simplifying this expression further:
dA/dx = 2 * sqrt(25 - x^2) + x^2 / sqrt(25 - x^2)
Setting this derivative equal to zero to find the critical points:
2 * sqrt(25 - x^2) + x^2 / sqrt(25 - x^2) = 0
Multiplying both sides of the equation by sqrt(25 - x^2):
2(25 - x^2) + x^2 = 0
Expanding and simplifying:
50 - 2x^2 + x^2 = 0
2x^2 = 50
x^2 = 25
Taking the square root of both sides and considering only the positive square root:
x = 5
Now that we have found the critical point x = 5, we can substitute this value back into the equation for the area to find the corresponding value of y:
A = 2x * sqrt(25 - x^2)
A = 2 * 5 * sqrt(25 - 5^2)
A = 10 * sqrt(25 - 25)
A = 0
The area of the rectangle at the critical point x = 5 is zero, which does not make sense in this context. Therefore, we need to consider the endpoints of the interval as well.
Since the rectangle is bounded by the x-axis and the semicircle, the possible values for x are in the interval [-5, 5].
First, let's consider the value of x = -5:
A = 2 * (-5) * sqrt(25 - (-5)^2)
A = -10 * sqrt(25 - 25)
A = -10 * sqrt(0)
A = 0
Next, let's consider the value of x = 5:
A = 2 * 5 * sqrt(25 - 5^2)
A = 10 * sqrt(25 - 25)
A = 0
So, at both endpoints of the interval, the area of the rectangle is zero.
Therefore, there is no maximum area for the rectangle within the given domain.