What is the largest rectangle that can be inscribed in the first quadrant of the ellipse 9x^2+16y^2=144?
base along x-axis, height along y-axis, bottom left vertex at (0,0).
let the top right vertex be (x,y)
from equation
y = (1/4)√(144 - 9x^2)
Area = xy
= x(1/4)√(144 - 9x^2)
differentiate using the product rule, set the derivative equal to zero and solve for x
I got x = 4√3/3 for a max area of 384√3/3
but check my arithmetic.
To find the largest rectangle that can be inscribed in the first quadrant of the ellipse, we need to maximize the area of the rectangle.
Let's start by expressing the equation of the ellipse in terms of y:
9x^2 + 16y^2 = 144
Dividing both sides of the equation by 144:
x^2/16 + y^2/9 = 1
Now let's consider a rectangle with its sides parallel to the x and y axes, where the bottom left corner of the rectangle is at the origin (0,0).
Let the length of the rectangle along the x-axis be 2a, and the height of the rectangle along the y-axis be 2b. So, the coordinates of the top right corner of the rectangle are (a, b).
We need to find the maximum values of a and b such that the rectangle is still inside the ellipse.
The equation of the ellipse can be rearranged as follows:
x^2/16 = 1 - y^2/9
This equation describes a horizontal hyperbola. Since we're looking for the largest rectangle inscribed in the first quadrant, the rectangle's vertices need to lie on this hyperbola.
The distance from the origin to the rectangle's top right corner (a, b) is given by the equation:
x^2 = a^2 = 16 - y^2
Substituting this into the equation for the ellipse:
16 - y^2 + 16y^2/9 = 144
Multiplying the equation by 9 to clear the fraction:
9(16) - 9y^2 + 16y^2 = 1296
144 - 9y^2 + 16y^2 = 1296
7y^2 = 1296 - 144
7y^2 = 1152
y^2 = 1152/7
y ≈ ±11.14
Since we only consider the positive value for y in the first quadrant, y ≈ 11.14.
Substituting this value back into the equation for the ellipse:
a^2 = 16 - y^2 ≈ 16 - 124.09 ≈ -108.09
Since we can't have a negative value for the length of the rectangle, this means that a ≈ 0.
Therefore, the largest rectangle that can be inscribed in the first quadrant of the ellipse has a length of 0 along the x-axis and a height of approximately 11.14 along the y-axis.
To find the largest rectangle that can be inscribed in the first quadrant of the ellipse, we can use optimization techniques. First, let's visualize the situation and understand the problem.
The equation of the ellipse is given as 9x^2 + 16y^2 = 144, which represents an ellipse centered at the origin (0,0) with a major axis along the x-axis and a minor axis along the y-axis. The goal is to find the largest rectangle that can fit entirely within the ellipse and is aligned with the x and y axes.
To solve this problem, we need to maximize the area of the rectangle. Let's assume the sides of the rectangle have lengths 2a and 2b, where a represents the distance from the origin to the side of the rectangle parallel to the x-axis, and b represents the distance to the side of the rectangle parallel to the y-axis.
We want to maximize the area A = 2a * 2b = 4ab.
Now, let's use the given equation of the ellipse to express the value of b in terms of a.
We have, 9x^2 + 16y^2 = 144. Rearranging, we get y^2 = (144 - 9x^2) / 16.
Since the rectangle is inscribed in the first quadrant, both x and y are positive. Therefore, the maximum value of y would be the value of y on the ellipse.
Substituting y^2 = (144 - 9x^2) / 16, we get:
b = y = sqrt((144 - 9x^2) / 16).
Now, we need to express the area A in terms of a only. We substitute b in A = 4ab:
A = 4a * sqrt((144 - 9x^2) / 16) = 2a * sqrt(144 - 9x^2).
To maximize A, we need to find the critical points of A. This can be done by finding the derivative of A with respect to x, setting it equal to zero, and solving for x.
dA/dx = 0
differentiating both sides with respect to x, we get,
2 * sqrt(144 - 9x^2) + 2a * (-9x / sqrt(144 - 9x^2)) = 0.
Simplifying this equation, we get:
2 * sqrt(144 - 9x^2) - 18ax / sqrt(144 - 9x^2) = 0.
Multiply both sides by sqrt(144 - 9x^2), we get:
2(144 - 9x^2) - 18ax = 0.
Expanding and rearranging the terms, we obtain:
288 - 18x^2 - 18ax = 0.
Now, let's solve this equation for x.