Math Problem Statement

Find the dimensions of the rectangle inscribed in the ellipse with the equation (x^2/25) + (y^2/16) = 1.

Solution

To solve the problem of finding the dimensions of the rectangle inscribed in the ellipse with the given equation:

x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1

Steps:

  1. Understand the ellipse equation:

    • The ellipse is centered at the origin (0,0)(0, 0) with semi-major axis a=5a = 5 (horizontal) and semi-minor axis b=4b = 4 (vertical).
  2. Key idea for an inscribed rectangle:

    • The vertices of the rectangle inscribed in the ellipse will lie on the ellipse.
    • Let the vertices of the rectangle be (x,y)(x, y), (x,y)(-x, y), (x,y)(x, -y), and (x,y)(-x, -y).
    • The rectangle's area is given by A=4xyA = 4xy, and we want to find xx and yy under the ellipse's constraint.
  3. Ellipse equation constraint:

    • The points (x,y)(x, y) must satisfy the ellipse equation: x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1
  4. Optimize the area:

    • Express yy in terms of xx using the ellipse equation: y2=16(1x225)y^2 = 16\left(1 - \frac{x^2}{25}\right) y=±41x225y = \pm 4\sqrt{1 - \frac{x^2}{25}}
    • The rectangle's area is: A=4xy=4x41x225A = 4xy = 4x\cdot 4\sqrt{1 - \frac{x^2}{25}} A=16x1x225A = 16x\sqrt{1 - \frac{x^2}{25}}
  5. Find xx that maximizes the area:

    • Differentiate A(x)A(x) with respect to xx and set the derivative to zero: dAdx=161x225+16x(2x251x225)\frac{dA}{dx} = 16\sqrt{1 - \frac{x^2}{25}} + 16x\left(-\frac{2x}{25\sqrt{1 - \frac{x^2}{25}}}\right) Simplify: dAdx=161x22532x2251x225\frac{dA}{dx} = 16\sqrt{1 - \frac{x^2}{25}} - \frac{32x^2}{25\sqrt{1 - \frac{x^2}{25}}} Set dAdx=0\frac{dA}{dx} = 0: 161x225=32x2251x22516\sqrt{1 - \frac{x^2}{25}} = \frac{32x^2}{25\sqrt{1 - \frac{x^2}{25}}} Multiply through by 1x225\sqrt{1 - \frac{x^2}{25}}: 16(1x225)=32x22516\left(1 - \frac{x^2}{25}\right) = \frac{32x^2}{25} Expand and solve: 1616x225=32x22516 - \frac{16x^2}{25} = \frac{32x^2}{25} 16=48x22516 = \frac{48x^2}{25} x2=40048=253x^2 = \frac{400}{48} = \frac{25}{3} x=±253=±53x = \pm\sqrt{\frac{25}{3}} = \pm\frac{5}{\sqrt{3}}
  6. Find corresponding yy:

    • Substitute x=53x = \frac{5}{\sqrt{3}} into the ellipse equation: y2=16(1(53)225)y^2 = 16\left(1 - \frac{\left(\frac{5}{\sqrt{3}}\right)^2}{25}\right) y2=16(125325)y^2 = 16\left(1 - \frac{\frac{25}{3}}{25}\right) y2=16(113)=1623=323y^2 = 16\left(1 - \frac{1}{3}\right) = 16\cdot\frac{2}{3} = \frac{32}{3} y=±423y = \pm\frac{4\sqrt{2}}{\sqrt{3}}
  7. Dimensions of the rectangle:

    • The rectangle has dimensions: 2x=253=1032x = 2\cdot\frac{5}{\sqrt{3}} = \frac{10}{\sqrt{3}} 2y=2423=8232y = 2\cdot\frac{4\sqrt{2}}{\sqrt{3}} = \frac{8\sqrt{2}}{\sqrt{3}}

Final Answer:

The dimensions of the rectangle are: 103×823\frac{10}{\sqrt{3}} \times \frac{8\sqrt{2}}{\sqrt{3}}

Would you like a detailed explanation for any step?

Here are 5 related questions you may find helpful:

  1. What is the maximum area of the rectangle inscribed in the ellipse?
  2. How can Lagrange multipliers be applied to solve this problem?
  3. How do the dimensions of the rectangle relate to the ellipse’s semi-major and semi-minor axes?
  4. What happens if the ellipse is tilted (rotated)?
  5. Can this approach be generalized to ellipses with arbitrary semi-axis lengths?

Tip: Always consider the symmetry of ellipses when solving geometric problems—it simplifies calculations!

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Math Problem Analysis

Mathematical Concepts

Ellipse Geometry
Optimization
Area Maximization

Formulas

Equation of ellipse: (x^2/a^2) + (y^2/b^2) = 1
Area of rectangle: A = 4xy
Derivative of a function for optimization

Theorems

Symmetry of Ellipse
Optimization via Differentiation

Suitable Grade Level

Grades 11-12