Find the dimensions of a rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse given by x^2/169 + y^2/49 = 1.
What would be the primary equation?
using symmetry, if one corner of the rectangle is at (h,k), the area is
a = 2h*2k = 4hk
Now, we know that k = 7√(1-h^2/169)
a = 4h*7√(1-h^2/169)
= 28/13 h√(169-h^2)
da/dh = 28/13 (2h^2-169)/√(169-h^2)
so, max area occurs when h = 13/√2
max area is 4 √13/2 7/√2 = √91
As expected, the corners are halfway to the ends of the ellipse.
hmmm. typo
max a = 4 13/√2 7/√2 = 91/2
To find the dimensions of the rectangle of maximum area inscribed in the given ellipse, we need to maximize the area of the rectangle.
Let the length of the rectangle be 2a and the width be 2b. Since the sides of the rectangle are parallel to the coordinate axes, we can write the equations as follows:
x = ± a
y = ± b
The area of the rectangle, A, is given by A = (2a) x (2b) = 4ab.
Now, we need to express one variable in terms of the other using the equation of the ellipse:
x^2/169 + y^2/49 = 1
Rearranging the equation, we have:
x^2/169 = 1 - y^2/49
x^2 = 169 - 169y^2/49
Dividing through by 169, we get:
x^2/169 = 1 - y^2/49
x^2 = 169 - (169/49)y^2
Plugging in x = ± a and y = ± b from the equations of the rectangle, we have:
a^2 = 169 - (169/49)b^2
Simplifying, we get:
(49/169)a^2 + b^2 = 1
This is the primary equation that relates the dimensions a and b of the rectangle to the ellipse equation.
To find the dimensions of the rectangle of maximum area inscribed in the given ellipse, we need to determine the equation that represents the area of the rectangle. Let's proceed step by step.
Step 1: Understand the problem
We are given an ellipse with the equation x^2/169 + y^2/49 = 1. We need to find the dimensions of a rectangle with sides parallel to the coordinate axes that can be inscribed in this ellipse. The rectangle will have maximum area.
Step 2: Determine the equation for the area of the rectangle
Let the length of the rectangle along the x-axis be 2a and the length along the y-axis be 2b. The area of a rectangle is given by the product of its length and width, so the area of this rectangle is A = (2a)(2b) = 4ab.
Step 3: Establish constraints
Since the sides of the rectangle are parallel to the coordinate axes, the coordinates of the vertices of the rectangle will be (±a, ±b).
Step 4: Relate the rectangle's vertices to the ellipse equation
The vertices (±a, ±b) of the rectangle lie on the ellipse given by the equation x^2/169 + y^2/49 = 1. We can substitute these coordinates into the equation to get: (a^2/169) + (b^2/49) = 1.
Step 5: Solve for one variable in terms of the other
Rearranging the equation from Step 4, we have:
(a^2/169) = 1 - (b^2/49)
a^2 = 169 - (169/49)b^2
a^2 = 169(1 - (b^2/49))
a^2 = 169 - (169/49)b^2
Step 6: Determine the maximum area
We want to maximize the area A = 4ab. To do so, we need to find the values of a and b which will maximize this product. Since we have an equation for a^2 in terms of b^2, we can substitute it into the equation for A to get an equation in terms of b only. Then, we can differentiate this equation with respect to b, set it equal to zero to find critical points, and determine the maximum value.
Once we find the value of b, we can substitute it back into the equation for a^2 to find a. Finally, we have the dimensions of the rectangle of maximum area.
The primary equation will be the equation for A (the area of the rectangle) in terms of b, after differentiating it and setting it equal to zero.