Use the method of Lagrange multipliers to find the area of the largest rectangle (centered at the origin) which can be inscribed in the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \).
We want to maximize the area of the rectangle, which is given by the product of its width and height. Let the width and height of the rectangle be \(2x\) and \(2y\) respectively. So the area of the rectangle is \(4xy\).
We want to maximize this area subject to the constraint that the rectangle is inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).
We need to find the critical points of the function \(f(x,y) = 4xy\) subject to the constraint \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\). We can use the method of Lagrange multipliers to solve this problem.
Let \(g(x,y) = \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1\) be the constraint equation.
The Lagrangian function is given by \(L(x,y,\lambda)=f(x,y)-\lambda g(x,y)\). We can write this as:
\(L(x,y,\lambda) = 4xy - \lambda \left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1\right)\)
To find the critical points, we need to find the partial derivatives of \(L\) with respect to \(x\), \(y\), and \(\lambda\), and set them equal to zero:
\(\frac{{\partial L}}{{\partial x}} = 4y - \frac{{2x}}{{a^{2}}}\lambda = 0 \Rightarrow 2y = \frac{{x}}{{a^{2}}}\lambda\)
\(\frac{{\partial L}}{{\partial y}} = 4x - \frac{{2y}}{{b^{2}}}\lambda = 0 \Rightarrow 2x = \frac{{y}}{{b^{2}}}\lambda\)
\(\frac{{\partial L}}{{\partial \lambda}} = \frac{{x^{2}}}{a^{2}} + \frac{{y^{2}}}{b^{2}} - 1 = 0\)
Solving these equations simultaneously, we get:
\(\frac{{x}}{{a^{2}}}\lambda = \frac{{y}}{{b^{2}}}\lambda \Rightarrow x = \frac{{a^{2}}}{b^{2}}y\)
Substituting this into the constraint equation, we get:
\(\frac{{\left(\frac{{a^{2}}}{b^{2}}y\right)^{2}}}{a^{2}} + \frac{{y^{2}}}{b^{2}} = 1\)
Simplifying, we get:
\(\frac{{a^{4}y^{2}}}{{b^{4}}} + \frac{{y^{2}}}{b^{2}} = 1\)
Multiplying through by \(b^{4}\), we get:
\(a^{4}y^{2} + b^{2}y^{2} = b^{4}\)
Factoring out \(y^{2}\), we get:
\(y^{2}(a^{4} + b^{2}) = b^{4}\)
Dividing through by \(a^{4} + b^{2}\), we get:
\(y^{2} = \frac{{b^{4}}}{{a^{4} + b^{2}}}\)
Taking square roots, we get:
\(y = \frac{{b^{2}}}{{\sqrt{a^{4} + b^{2}}}}\)
Substituting this value of \(y\) into the equation \(x = \frac{{a^{2}}}{b^{2}}y\), we get:
\(x = \frac{{a^{2}}}{b^{2}}\frac{{b^{2}}}{{\sqrt{a^{4} + b^{2}}}} = \frac{{a^{2}}}{{\sqrt{a^{4} + b^{2}}}}\)
So the dimensions of the rectangle that maximizes the area are:
\(2x = \frac{{2a^{2}}}{{\sqrt{a^{4} + b^{2}}}}\) and \(2y = \frac{{2b^{2}}}{{\sqrt{a^{4} + b^{2}}}}\)
Hence, the area of the largest rectangle that can be inscribed in the ellipse is given by the product of the width and height:
\(\text{{Area}} = \frac{{2a^{2}}}{{\sqrt{a^{4} + b^{2}}}} \cdot \frac{{2b^{2}}}{{\sqrt{a^{4} + b^{2}}}} = \frac{{4a^{2}b^{2}}}{{a^{4} + b^{2}}}\)
To find the area of the largest rectangle inscribed in the ellipse \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \) centered at the origin, we can use the method of Lagrange multipliers.
Let's set up the problem using Lagrange multipliers. The objective function is the area of the rectangle, given by \( A = 4xy \), where \( x \) and \( y \) are the dimensions of the rectangle.
The constraint is the equation of the ellipse \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \).
We can introduce a Lagrange multiplier \( \lambda \) and solve the following system of equations:
1. \( \frac{\partial}{\partial x}(4xy) = \lambda \frac{\partial}{\partial x}\left(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1\right) \)
2. \( \frac{\partial}{\partial y}(4xy) = \lambda \frac{\partial}{\partial y}\left(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1\right) \)
3. \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \)
Taking the partial derivatives, we get:
1. \( 4y = \frac{2\lambda x}{a^{2}} \)
2. \( 4x = \frac{2\lambda y}{b^{2}} \)
From equations 1 and 2, we can solve for \( \lambda \):
\( \lambda = \frac{2ab^{2}y}{x} = \frac{2a^{2}bx}{y} \)
Substituting \( \lambda \) back into equations 1 and 2, we get:
1. \( 2y^{2} = ax^{2} \)
2. \( 2x^{2} = by^{2} \)
Simplifying the equations further, we have:
1. \( y^{2} = \frac{a}{2}x^{2} \)
2. \( x^{2} = \frac{b}{2}y^{2} \)
Multiplying equations 1 and 2 together, we obtain:
\( x^{2}y^{2} = \frac{ab}{4}x^{2}y^{2} \)
Dividing both sides by \( x^{2}y^{2} \) (assuming \( x \neq 0 \) and \( y \neq 0 \)), we get:
\( \frac{ab}{4} = 1 \)
Simplifying further, we find:
\( ab = 4 \)
Since we want to maximize the area of the rectangle, which is given by \( A = 4xy \), we need to find the maximum value of \( A \) subject to the constraint \( ab = 4 \).
To find the maximum, we can calculate \( A \) at the endpoints of the constraint. Since the rectangle is centered at the origin, \( x \) and \( y \) are positive. Therefore, the maximum occurs at \( a = 2 \) and \( b = 2 \).
Substituting \( a = 2 \) and \( b = 2 \) into the area formula \( A = 4xy \), we get:
\( A = 4 \cdot 2 \cdot 2 = 16 \)
Thus, the maximum area of the rectangle inscribed in the ellipse \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \) (centered at the origin) is 16 square units.