A rectangle is insribed between the parabolas y=4(x^2) and y=30-x^2. What is the maximum area of the rectangle?
Domain: [-3,3]
Range:[-2, 40]
THANK YOU!
First you have to find their intersection
which is an easy equation to solve
4x^2 = 30-x^2
x = ± √6
y = 24
so area = integral (30 - 5x^2) from -√6 to √6
or
= integral 2(30-5x^2) from 0 to √6
= 60x - (5/3)x^3 | from 0 to √6
= 60√6 - (5/3)6√6 - 0
= 50√6
To find the maximum area of the rectangle inscribed between the parabolas y=4(x^2) and y=30-x^2, we need to find the dimensions of the rectangle that will give us the largest possible area.
Let's start by visualizing the problem. The parabolas y=4(x^2) and y=30-x^2 are shown below within the given domain and range:
Graph:
-----------------------
| -------- |
| --- --- |
| -2 2 |
| | |
| ----------------- |
| | | |
| | Rectangle | |
| | | |
| ----------------- |
| | |
| | --- |
| --- --- |
| -------- |
-----------------------
Now, the rectangle is inscribed between these two parabolas. Let's find the points of intersection between the two parabolas to determine the length of the rectangle.
Setting the two equations equal to each other:
4(x^2) = 30 - x^2
Combining like terms:
5x^2 = 30
Dividing both sides by 5:
x^2 = 6
Now, taking the square root of both sides and considering that the rectangle is inscribed between x values -3 and 3, we get:
x = ±√6
Since we are interested in the length of the rectangle, we need to consider the positive value of x, so x = √6.
Now, we need to find the height of the rectangle. This can be calculated by finding the difference between the y values of the two parabolas at x = √6.
For y=4(x^2):
y = 4(√6)^2
Simplifying:
y = 4 * 6 = 24
For y=30-x^2:
y = 30 - (√6)^2
Simplifying:
y = 30 - 6 = 24
Notice that the y values are the same for both parabolas. This means that the rectangle formed will be a square.
The length and height of the rectangle are both equal to 24, so the area of the rectangle is:
Area = Length * Height = 24 * 24 = 576
Therefore, the maximum area of the rectangle inscribed between the two parabolas is 576 square units.