A rectangle is bounded by the x-axis and the semicircle y = √36 – x2, as shown in the figure below. Write the area A of the rectangle as a function of x, and determine the domain of the area function.
A =
all real numbers except x = 36
all real numbers except x = 6
0 < x < 6
0 < x < 36
all real numbers
I don't see any "figure below" , but I can surmise your figure shows a quarter circle in the first quadrant with a rectangle, whose base is along the x-axis, its height along the y-axis and it touches the quarter circle at (x,y)
Area = xy
= x√(36-x^2) or x(36-x^2)^(1/2)
for the domain, we have to make sure that the number under the √ does not become negative, so
0 < x < 6
Well, let's break it down! The area of a rectangle is given by length times width. In this case, the length of the rectangle is given by the distance between the x-axis and the semicircle, which can be represented by the expression y = √36 – x^2.
Since the rectangle is bounded by the x-axis, the width is simply given by 2x. Therefore, the area A of the rectangle can be written as:
A = (2x)(√36 – x^2)
Now, let's determine the domain of the area function. We need to find the values of x that make the expression (√36 – x^2) non-negative, as the length cannot be negative.
√36 – x^2 ≥ 0
Taking square root on both sides, we get:
√36 ≥ x^2
6 ≥ x^2
Taking square root on both sides again, we get:
√6 ≥ x
Therefore, the domain of the area function is 0 < x ≤ √6, or in interval notation, (0, √6].
So, the correct answer is 0 < x < 6. But hey, let's face it, what's a little math without a little clowning around? I've got a joke for you:
Why did the rectangle become an astronaut?
Because he wanted to go on a space "rectangle" mission!
To find the area of the rectangle, we need to determine the length and width of the rectangle.
The length of the rectangle is the distance between the x-axis and the semicircle. Since the semicircle is defined by the equation y = √(36 – x^2), the length can be calculated as the difference between the y-coordinate of the semicircle and the x-axis, which is √(36 - x^2) - 0 = √(36 - x^2).
The width of the rectangle is the distance between the two points where the semicircle intersects with the x-axis. To find these points, we set y = 0 in the equation y = √(36 - x^2):
0 = √(36 - x^2)
0 = 36 - x^2
x^2 = 36
x = ±6
So, the two points of intersection are x = -6 and x = 6. As the rectangle is bounded by the x-axis, the width is the distance between these two points, which is 6 - (-6) = 12.
Now, we can calculate the area of the rectangle by multiplying the length and width:
A = Length * Width
A = (√(36 - x^2)) * 12
A = 12√(36 - x^2)
The domain of the area function is determined by the restrictions on x-values that make sense in the context of the problem. Since the semicircle equation has a square root, the expression under the square root must be non-negative:
36 - x^2 ≥ 0
x^2 ≤ 36
Taking the square root of both sides, we get:
-6 ≤ x ≤ 6
Therefore, the domain of the area function is -6 ≤ x ≤ 6. However, since we want to find the area of the rectangle as a function of x, we can further refine the domain to exclude the endpoints where x = -6 and x = 6. This gives us the final domain:
-6 < x < 6
To write the area A of the rectangle as a function of x, we need to find the length and width of the rectangle.
The length of the rectangle is the difference between the x-coordinates of the two points where the semicircle intersects the x-axis. To find these points, we set y = 0 in the equation of the semicircle:
0 = √36 - x^2
Solving for x, we have:
x^2 = 36
x = ±√36
x = ±6
Since the rectangle is bounded by the x-axis, the length of the rectangle is 2 times the absolute value of the x-coordinate:
length = 2|x| = 2(6) = 12
The width of the rectangle is the y-coordinate of the point on the semicircle that has the same x-coordinate as the rectangle. Substituting x into the equation of the semicircle, we have:
y = √36 - x^2
y = √36 - 36
y = 0
Therefore, the width of the rectangle is 0.
Now we can calculate the area A of the rectangle:
A = length × width
A = 12 × 0
A = 0
So, the area A of the rectangle is 0 for all values of x within the interval [0, 6].
Therefore, the domain of the area function is 0 < x < 6. Thus, the answer is 0 < x < 6.