A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=12−x2. What are the dimensions of such a rectangle with the greatest possible area?
let the points of contact be (x,y) and (-x,y)
let the area of the rectangle be A
A = 2xy
= 2x(12-x^2) = 24x - 2x^3
dA/dx = 24 -6x^2
= 0 for a max of A
6x^2 = 24
x^2 = 4
x = ± 2
y = 12 - 4 = 8
the rectangle will be 4 units for the base and the height will be 8
(touching at the points (2,8) and (-2,8)
To find the dimensions of the rectangle with the greatest possible area, we need to analyze the problem step by step.
Step 1: Understand the problem
We are given a rectangle that is inscribed with its base on the x-axis. This means that the base of the rectangle lies on the x-axis, and the rectangle's upper corners touch the parabola y = 12 - x^2. We need to find the dimensions of this rectangle that maximize its area.
Step 2: Define variables
Let's define the dimensions of the rectangle as follows:
- Length of the rectangle: l
- Width of the rectangle: w
Step 3: Express the problem mathematically
The area of a rectangle is given by the formula: Area = length × width.
So, the area of our rectangle can be expressed as: Area = l × w.
Step 4: Determine constraints
Since the rectangle is inscribed on the parabola y = 12 - x^2, its upper corners lie on this curve. We can use this information to set up the constraints for our problem.
The x-coordinate of the upper corners of the rectangle will be the same as the x-coordinate of the point on the parabola. This can be expressed as:
x = l/2
The y-coordinate of the upper corners of the rectangle will be the y-coordinate of the point on the parabola. This can be expressed as:
y = 12 - x^2
Step 5: Express the area in terms of a single variable
Since we have set up the constraints, we can now express the area of the rectangle in terms of a single variable, either l or w. Let's express it in terms of l, using the constraints we derived in Step 4:
Area = l × w
= l × (12 - (l/2)^2)
= l × (12 - l^2/4)
= 12l - (l^3)/4
Step 6: Maximize the area
To find the dimensions of the rectangle with the greatest possible area, we need to maximize the Area expression from Step 5. We can find the maximum by finding the critical points of the function, which occur when the derivative is equal to zero.
Differentiating the Area expression with respect to l:
d(Area)/dl = 12 - (3l^2)/4
Setting the derivative equal to zero:
12 - (3l^2)/4 = 0
Simplifying and solving for l^2:
3l^2 = 48
l^2 = 48/3
l^2 = 16
l = ±4
Since a length cannot be negative, l = 4.
Step 7: Find the width and the dimensions of the rectangle
Using the derived length l, we can find the width of the rectangle using the constraint x = l/2:
x = l/2
= 4/2
= 2
To find the y-coordinate, we substitute x = 2 into the equation of the parabola y = 12 - x^2:
y = 12 - (2)^2
= 12 - 4
= 8
Therefore, the dimensions of the rectangle with the greatest possible area are:
Length = 4 units
Width = 2 units