A rectangle has its base on the x-axis and its upper two vertices on the parabola
y=12−x2 .
What is the largest area that the rectangle can have
let the points of contact be (x,y) and (-x,y)
so the base is 2x and the height is y
but y=12-x^2
so the
Area = 2xy
= 2x(12-x^2)
= 24x - 2x^3
d(Area)/dx = 24 - 6x^2
= 0 for a max Area
solve 2x^3 - 24 = 0
sub that back into the Area = 2x(12-x^2)
To find the largest area that the rectangle can have, we need to consider that the rectangle's vertices lie on the parabola y = 12 - x^2 and the base of the rectangle is on the x-axis.
Let's visualize the problem to understand it better.
1. First, let's draw the coordinate plane.
2. Next, plot the graph of the parabola y = 12 - x^2.
3. The base of the rectangle lies on the x-axis, so it will have a length equal to the x-coordinate of the rectangle's vertices. Let's call this length "2x" since it extends from -x to x on the x-axis.
4. The height of the rectangle is the vertical distance between the x-axis and the parabola. This distance is given by y = 12 - x^2.
5. Now, let's express the area of the rectangle in terms of x. The area of a rectangle is given by length times width. In this case, the width is 2x, and the height is y = 12 - x^2. So the area A can be expressed as: A = 2x * (12 - x^2).
6. To find the largest area, we need to maximize the area function A(x).
To find the largest area, we will use calculus. We can take the derivative of the area function A(x) with respect to x, and then set it equal to zero to solve for the critical points.
7. Taking the derivative of A(x) = 2x * (12 - x^2) with respect to x gives us: A'(x) = 2(12 - x^2) - 4x^2.
8. Setting A'(x) equal to zero and solving for x gives us: 2(12 - x^2) - 4x^2 = 0.
Simplifying, we get: 24 - 2x^2 - 4x^2 = 0.
Combining like terms: 24 - 6x^2 = 0.
9. Solving the quadratic equation, we get: 6x^2 = 24.
Dividing both sides by 6, we find: x^2 = 4.
Taking the square root of both sides, we get: x = ± 2.
10. Since we are looking for a positive length, we can ignore the negative value of x. Therefore, x = 2.
11. Now that we have the value of x, we can substitute it back into the area function A(x) = 2x * (12 - x^2) to find the largest area.
A(2) = 2(2) * (12 - 2^2) = 4 * (12 - 4) = 4 * 8 = 32.
Therefore, the largest area that the rectangle can have is 32 square units.
To find the largest area that the rectangle can have, we need to maximize the area equation. The area of a rectangle is given by length multiplied by width.
Let's first consider the rectangle's base, which lies on the x-axis. Given that the base of the rectangle is on the x-axis, it means that the width of the rectangle is equal to the distance between the x-coordinates of the two upper vertices of the rectangle.
To find the width, we need to determine the x-coordinates of the upper vertices. Since the parabola equation is given as y = 12 - x^2, we can set y = 0 and solve for x to find the x-intercepts:
0 = 12 - x^2
x^2 = 12
x = ±√12
x = ±2√3
Therefore, the x-coordinates of the upper vertices of the rectangle are -2√3 and 2√3.
Now let's calculate the width:
width = distance between x-coordinates of the upper vertices
= 2√3 - (-2√3)
= 4√3
To maximize the area, we need to find the length of the rectangle. The length is determined by the y-coordinate difference between the upper and lower vertices. In this case, the rectangle lies on the parabola, so the length is equal to the y-coordinate difference between the lower vertex (0, 12) and one of the upper vertices.
length = y-coordinate of the upper vertex - y-coordinate of the lower vertex
= (12 - x^2) - 12
= -x^2
Since we want to maximize the area, we need to find the maximum value of -x^2. The maximum value of a negative quadratic occurs at the vertex.
The vertex of the parabola y = -x^2 occurs at x = 0. Therefore, the maximum value of -x^2 is 0.
Now let's calculate the area:
area = length * width
= -x^2 * 4√3
= -0 * 4√3
= 0
The largest area that the rectangle can have is 0.