Problem 28 - Calculus Early Transcendentals

Question

Applications of Lagrange multipliers Use Lagrange multipliers in the following problems. When the domain of the objective function is unbounded or open, explain why you have found an absolute maximum or minimum value. Maximum area rectangle in an ellipse Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse \(4 x^{2}+16 y^{2}=16\).

Step-by-Step Solution

Answer

The dimensions of the rectangle with maximum area inscribed in the given ellipse are \(2x = \sqrt{17}\) and \(2y = 1\).

Step 1: Set up the Lagrange Multiplier equation

The Lagrange Multiplier equation is given by the following expression: \(\nabla A(x, y) = \lambda \nabla g(x, y)\) where \(\nabla\) denotes the gradient operator, \(A(x,y)\) is our objective function (area), \(\lambda\) is the Lagrange multiplier, and \(g(x,y)\) is our constraint function (i.e., the ellipse equation).

Step 2: Find the gradients of A(x, y) and g(x, y)

Next, we need to find the gradients of the objective and constraint functions: \(\nabla A(x, y) = \left(\frac{\partial}{\partial x}(4xy) , \frac{\partial}{\partial y}(4xy) \right) = (4y, 4x)\) \(\nabla g(x, y) = \left(\frac{\partial}{\partial x}(4x^2 + 16y^2 - 16) , \frac{\partial}{\partial y}(4x^2 + 16y^2 - 16) \right) = (8x, 32y)\)

Step 3: Set the Lagrange Multiplier equation

Now, we can set up the Lagrange Multiplier equation as follows: \((4y, 4x) = \lambda (8x, 32y)\) This equation gives us two more equations: \(4y = 8x\lambda \hspace{15mm}(1)\) \(4x = 32y\lambda \hspace{15mm}(2)\)

Step 4: Solve the system of equations

Now, we have to solve the system of equations (1), (2), and the constraint \(4x^2 + 16y^2 = 16\) simultaneously: Divide equation \((1)\) by \(4\): \(y = 2x\lambda\) Square both sides: \(y^2 = 4x^2\lambda^2\) Now, multiply equation \((2)\) by \(8\): \(32x = 256y\lambda\) Then divide by \(32\): \(x = 8y\lambda\) Square both sides: \(x^2 = 64y^2\lambda^2\) Now, plug \(x^2\) and \(y^2\) into the equation of the ellipse: \(4(64y^2\lambda^2) + 16(4x^2\lambda^2) = 16\) Simplify: \(256y^2\lambda^2 + 64x^2\lambda^2 = 16\) Factor out \(64\lambda^2\): \(64\lambda^2(x^2 + 4y^2) = 16\) Divide by \(64\lambda^2\): \(x^2 + 4y^2 = \frac{1}{4\lambda^2}\) Now, plug in \(y = 2x\lambda\): \(x^2 + 16x^2\lambda^2 = \frac{1}{4\lambda^2}\) Combine like terms: \(17x^2\lambda^2 = \frac{1}{4\lambda^2}\) Divide by \(17\lambda^2\): \(x^2 = \frac{1}{68\lambda^4}\)

Step 5: Find x and y coordinates

Now, either we have \(\lambda = 0\) and x = 0, or \(\lambda \neq 0\) and we can obtain a solution for x. If \(\lambda = 0\), \(4y = 0\) from equation (1), and x = 0, y = 0, which doesn't give us the dimensions of the rectangle. If \(\lambda \neq 0\), divide by \(\lambda^4\): \(x^2 = \frac{1}{68\lambda^4}\) Denote \(z = \frac{1}{\lambda^2}\): \(x^2 = \frac{z^2}{68}\) Plug this value back into the ellipse equation: \(4\left(\frac{z^2}{68}\right) + 16y^2 = 16\) Solve for \(y^2\): \(y^2 = \frac{1}{4}\left(4 - \frac{z^2}{17}\right)\) Now, we can find x and y values: \(x = \frac{z}{\sqrt{68}}\) \(y = \frac{1}{2}\sqrt{4 - \frac{z^2}{17}}\) Use the equation \(y = 2x\lambda\), plug \(x = \frac{z}{\sqrt{68}}\) and solve for \(z\): \(z = 2\sqrt{17}\) Plug this z-value into the expression for x and y: \(x = \frac{2\sqrt{17}}{\sqrt{68}} = \frac{\sqrt{17}}{2}\) \(y = \frac{1}{2}\sqrt{4 - \frac{(2\sqrt{17})^2}{17}} = \frac{1}{2}\) Thus, the dimensions of the rectangle with maximum area inscribed in the given ellipse are \(2x = \sqrt{17}\) and \(2y = 1\).

Key Concepts

Constrained Optimization

Constrained optimization is a crucial mathematical process where we find the maximum or minimum values of a function subject to certain limitations or constraints. In simpler terms, it's like trying to find the optimal solution (say maximum happiness) but under practical limits (like budget or time constraints). In the given exercise, we are dealing with a constrained optimization problem where we are trying to maximize the area of a rectangle. The constraint here is the rectangle must fit inside an ellipse.

• The objective function is what we are trying to maximize or minimize. In our case, it's the area of the rectangle.
• The constraint is what limits the values that the objective function can take. For us, this is the ellipse equation.

By applying Lagrange multipliers, we can efficiently tackle this sort of problem even when the solution isn't straightforward.

Ellipse

An ellipse is a fundamental geometric shape that looks like a squashed circle. It's a set of all points for which the sum of the distances from two fixed points (foci) is constant. In mathematical terms, it gives us a constraint to work with when identifying the area of a rectangle inside it.
In our problem, the equation of the ellipse is given by:
\[4x^2 + 16y^2 = 16\]

• Here, the terms in the equation help define the shape and size of the ellipse on the coordinate plane.
• This provides a boundary within which our rectangle must be inscribed to maximize its area without breaking out of the ellipse.

Understanding the shape of this boundary is key to solving constrained optimization problems involving ellipses.

Gradient

A gradient is like a pointer showing the direction of greatest increase of a function. It's a vector consisting of partial derivatives of the function with respect to its variables.
In our problem, gradients help us understand how to adjust the dimensions of the rectangle to maximize the area while staying within the boundary of the ellipse.

• The gradient of the objective function (here, the area of the rectangle) indicates how the area changes with changes in rectangle dimensions.
• The gradient of the constraint function (the ellipse equation) shows how changes in the ellipse's edge affect our rectangle.
• With Lagrange multipliers, we set these gradients proportional, meaning we adjust dimensions until any change would mean violating the ellipse constraint.

This concept is especially useful in multidimensional optimization where direct solutions aren't always feasible.

Maximum Area Rectangle

Finding a rectangle with maximum area that is inscribed within a specific shape involves understanding both the shape's boundary and the rectangle's geometry.
In this case:

• The rectangle must have sides parallel to the coordinate axes, simplifying our calculations.
• We use the algebraic expressions derived from setting gradients equal to determine side lengths.
• By solving the given system of equations, the sides of the rectangle turn out to be \(2x = \sqrt{17}\) and \(2y = 1\).

Thus, the maximum area we can achieve for such a rectangle requires precise calculation and a clear understanding of both constraint and geometry. These allow us to draw the boundaries and maximize what is possible within them.