A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y= 1-x^2. What are the dimensions of such a rectangle with the greatest possible area?
To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area function.
Let's assume the length of the rectangle is 2x (base on the x-axis) and the height of the rectangle is y (distance from the x-axis to the upper corners on the parabola).
The area of the rectangle is given by the formula: A = length × height.
Substituting the given values, the area of the rectangle is: A = 2x × y.
Now, we need to express y in terms of x. Since the upper corners of the rectangle lie on the parabola y = 1 - x^2, we can substitute y with 1 - x^2 in the area formula:
A = 2x × (1 - x^2).
Next, we can expand the expression:
A = 2x - 2x^3.
To find the maximum value of this area function, we need to take its derivative with respect to x and set it equal to zero.
dA/dx = 2 - 6x^2 = 0.
Solving this equation for x, we get:
6x^2 = 2,
x^2 = 1/3.
Taking the square root of both sides:
x = ±√(1/3).
Since we are looking for a rectangle on the x-axis, we can ignore the negative value and focus on the positive value:
x = √(1/3).
Now, we can substitute this value of x back into the equation for y:
y = 1 - x^2,
y = 1 - (1/3),
y = 2/3.
Therefore, the dimensions of the rectangle with the greatest possible area are: length = 2(√(1/3)) and height = 2/3.
To find the dimensions of the rectangle with the greatest possible area, we need to consider the following:
1. The base of the rectangle lies on the x-axis.
2. The upper corners of the rectangle lie on the parabola y = 1 - x^2.
3. The area of a rectangle is given by the product of its length and width.
Let's proceed step by step:
Step 1: Determining the base of the rectangle
Since the base of the rectangle lies on the x-axis, the length of the rectangle will be the difference between the x-coordinates of the two upper corners of the rectangle. Let's say that the x-coordinate of the left corner is a, and the x-coordinate of the right corner is b. Therefore, the length of the rectangle is given by b - a.
Step 2: Determining the width of the rectangle
The width of the rectangle is the difference between the y-coordinates of the upper corners of the rectangle. Since the right corner lies on the parabola y = 1 - x^2, the y-coordinate of the right corner can be determined by substituting the x-coordinate of the right corner (which is b) into the equation. Therefore, the width of the rectangle is given by 1 - b^2.
Step 3: Determining the area of the rectangle
The area of the rectangle is given by the product of its length and width. Thus, the area A is given by A = (b - a)(1 - b^2).
Step 4: Maximizing the area
To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area function A = (b - a)(1 - b^2). We can do this by taking the derivative of A with respect to both variables (a and b) and equating the derivatives to zero to find the critical points.
Taking the derivative of A with respect to a, we get:
dA/da = - (1 - b^2)
Taking the derivative of A with respect to b, we get:
dA/db = (1 - a)(-2b) - (b - a)(-2b)
Equating the derivatives to zero, we have:
- (1 - b^2) = 0,
(1 - a)(-2b) - (b - a)(-2b) = 0
Solving these equations will give us the x-coordinates of the upper corners of the rectangle, which will determine its dimensions.
By solving the first equation, we find that b = ±1.
Substituting b = 1 into the second equation, we can solve for a:
(1 - a)(-2) - (1 - a)(-2) = 0
-2(1 - a) = 0
1 - a = 0
a = 1
Therefore, we have a rectangle with upper corners at (1, 0) and (-1, 0) on the parabola y = 1 - x^2. The dimensions of this rectangle are:
Length: b - a = 1 - (-1) = 2
Width: 1 - b^2 = 1 - 1^2 = 0
Hence, the rectangle with the greatest possible area has dimensions of length 2 and width 0.
let the point of contact of the top right corner with the parabola be (x,y)
then the base of the rectangle is 2x and its height is y
Area = 2xy
= 2x(1 - x^2)
= 2x - 2x^3
D(area)/dx = 2 = 6x^2
= 0 for a max of area
solve for x , then find y ...
very easy from here on