Find the rectangle with the maximum area inside the ellipse x^2/16 + y^2/4=1.
Thanks.
To find the rectangle with the maximum area inside the ellipse x^2/16 + y^2/4 = 1, we can use the concept of inscribed rectangles.
1. First, let's consider a rectangle that is aligned with the x and y-axis, with sides parallel to those axes. Let the sides of the rectangle be 2a and 2b, where a represents half the width of the rectangle, and b represents half the height.
2. Since the rectangle is inscribed inside the ellipse, its vertices will touch the ellipse. Let's find the coordinates of these vertices.
3. The coordinates of the vertices will be (+/- a, +/- b). Substituting these values into the equation of the ellipse, we get:
(a^2)/16 + (b^2)/4 = 1
4. Rearranging the equation, we have:
(a^2)/16 = 1 - (b^2)/4
5. Multiplying both sides of the equation by 16, we get:
a^2 = 16 - 4(b^2)
6. Taking the square root of both sides, we obtain:
a = sqrt(16 - 4(b^2))
7. The area of the rectangle is given by A = (2a)(2b) = 4ab. Substituting the value of a from the previous step, we get:
A = 4sqrt(16 - 4(b^2))b
8. To maximize the area, we need to find the value of b that maximizes this expression. To do that, we can take the derivative of the area function with respect to b and set it equal to zero.
dA/db = 4(sqrt(16 - 4b^2) - (4b^2/sqrt(16 - 4b^2)) = 0
9. Simplifying the equation, we get:
sqrt(16 - 4b^2) - b^2/sqrt(16 - 4b^2) = 0
10. Multiplying both sides by sqrt(16 - 4b^2), we obtain:
16 - 4b^2 - b^2 = 0
11. Combining like terms, we get:
5b^2 = 16
12. Solving for b, we have:
b^2 = 16/5
b = sqrt(16/5)
b = 2sqrt(5)/sqrt(5)
b = 2
13. Substituting the value of b into the equation of the ellipse, we can find the value of a:
(a^2)/16 + (2^2)/4 = 1
(a^2)/16 + 4/4 = 1
(a^2 + 16)/16 = 1
a^2 + 16 = 16
a^2 = 0
14. Since a^2 = 0, a = 0. This means that the width of the rectangle is zero, and the rectangle reduces to a line segment along the y-axis.
Therefore, the rectangle with maximum area inside the ellipse x^2/16 + y^2/4 = 1 is a line segment with one end at (0, 2) and the other end at (0, -2).
To find the rectangle with the maximum area inside the given ellipse, we need to maximize the area of the rectangle. Since the area of a rectangle is given by the product of its length and width, we need to find the dimensions that satisfy the conditions of being inside the ellipse and yield the maximum value for the area.
Let's assume the length of the rectangle is 2x and the width is 2y. This way, the rectangle is centered at the origin and each side of the rectangle is divided into two equal halves by the ellipse.
The equation of the ellipse is x^2/16 + y^2/4 = 1. We substitute x = 2x and y = 2y:
(2x)^2/16 + (2y)^2/4 = 1
4x^2/16 + 4y^2/4 = 1
x^2/4 + y^2 = 1
Now, we need to find the maximum values of x and y that satisfy this equation. Let's solve for y in terms of x:
y^2 = 1 - x^2/4
y = √(1 - x^2/4)
To find the maximum values of x and y, we need to find the extreme values of the square root function. Let's take the derivative of y with respect to x and find its critical points:
dy/dx = (1/2) * (-2x) / √(1 - x^2/4)
dy/dx = -x / √(1 - x^2/4)
To find the critical points, we set dy/dx = 0 and solve for x:
-x / √(1 - x^2/4) = 0
-x = 0
x = 0
So, x = 0 is a critical point.
To determine if it is a maximum or minimum, we need to check the second derivative. Let's take the second derivative of y:
d^2y/dx^2 = d/dx (-x / √(1 - x^2/4))
d^2y/dx^2 = (d/dx) (-x) * (1 - x^2/4)^(-1/2)
d^2y/dx^2 = -1 * (1 - x^2/4)^(-1/2) + x^2 * 1/2 * (1 - x^2/4)^(-3/2) * (-1/4) * (-2x)
d^2y/dx^2 = (1 - x^2/4)^(-1/2) - (x^2/2) / (1 - x^2/4)
Now, substitute x = 0 into this second derivative:
d^2y/dx^2 = (1 - 0^2/4)^(-1/2) - (0^2/2) / (1 - 0^2/4)
d^2y/dx^2 = 1 - 0 / 1
d^2y/dx^2 = 1
The second derivative is positive, which means the critical point x = 0 is a minimum. However, we are interested in finding the maximum values for x and y, so we need to consider the endpoints of the curve in the ellipse.
The endpoints occur when x = ±4 because the equation of the ellipse is x^2/16 + y^2/4 = 1. Plugging x = ±4 into the equation of the ellipse to solve for y:
(±4)^2/16 + y^2/4 = 1
16/16 + y^2/4 = 1
1 + y^2/4 = 1
y^2/4 = 0
y = 0
So, we have two endpoints: (4, 0) and (-4, 0), which represent the maximum values of x and y that satisfy the ellipse equation.
Therefore, the dimensions of the rectangle with maximum area that fits inside the given ellipse are 2(4) = 8 units for the length and 2(0) = 0 units for the width. This means the rectangle is degenerate and reduces to a line segment with an area of 0 units.
Hence, the rectangle with the maximum area inside the ellipse x^2/16 + y^2/4=1 is a line segment with an area of 0 units.
If the base of the rectangle has length 2x, and height y, the area is
a = 2xy = 2x√(1-x^2/16)(4)
= 2x√(16-x^2)
da/dx = 4(8-x^2)/√(16-x^2)
so, max area occurs at x=√8, and the rectangle is 2√8 by √2