Problem 27 - 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. Extreme distances to an ellipse Find the minimum and maximum distances between the ellipse \(x^{2}+x y+2 y^{2}=1\) and the origin.

Step-by-Step Solution

Answer

The minimum distance between the origin and the ellipse is \(\frac{\sqrt{2}}{\sqrt{5}}\), and there is no maximum distance.

Step 1: Define the objective function, constraint, and Lagrange function

Our objective function is the square of the distance to the origin: $$D^2=x^2+y^2$$The constraint is the equation of the ellipse: $$g(x, y) = x^2+xy+2y^2-1 = 0$$Now we can define the Lagrange function: $$L(x,y,\lambda)=x^2+y^2+\lambda(x^2+xy+2y^2-1)$$

Step 2: Find the critical points

To find the critical points, we'll take the partial derivatives of the Lagrange function with respect to x, y, and \(\lambda\), and set them equal to zero: $$\frac{\partial L}{\partial x}=2x+\lambda(2x+y)=0$$ $$\frac{\partial L}{\partial y}=2y+\lambda(x+4y)=0$$ $$\frac{\partial L}{\partial \lambda}=x^2+xy+2y^2-1=0$$

Step 3: Solve the system of equations

We have 3 equations and 3 unknowns (\(x\), \(y\), and \(\lambda\)). We can solve the system of equations by substitution or elimination. Let's use substitution. From the first equation, we get: $$\lambda=\frac{-2x}{2x+y}$$ Substitute this into the second equation: $$2y+\frac{-2x}{2x+y}(x+4y)=0$$ Multiply both sides by \((2x+y)\) to eliminate the fraction: $$2y(2x+y)-2x(x+4y)=0$$ Expand and simplify: $$4xy+2y^2-2x^2-8xy=0$$ Combining terms and rearranging, we get: $$2x^2-4xy+2y^2=0$$ Divide by 2: $$x^2-2xy+y^2=0$$ Which simplifies to: $$(x-y)^2=0$$ Therefore, \(x=y\). We can substitute y for x in the constraint equation: $$g(x, y) = x^2+x(x)+2(x)^2-1 = 0$$ Solve for x: $$4x^2+x^2-1=0$$ $$5x^2=1$$ $$x^2=\frac{1}{5}$$ $$x=\pm\frac{1}{\sqrt{5}}$$ Since \(y=x\), we have two critical points: \((-\frac{1}{\sqrt{5}},-\frac{1}{\sqrt{5}})\) and \((\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}})\).

Step 4: Evaluate the objective function at the critical points

Now we will evaluate the objective function \(D^2=x^2+y^2\) at the critical points: At \((-\frac{1}{\sqrt{5}},-\frac{1}{\sqrt{5}})\): $$D^2=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$$ At \((\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}})\): $$D^2=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$$ Since the domain of the objective function is unbounded, we will look for an increasing gradient moving away from the boundary to confirm that these points represent the minimum distance. The gradient of the ellipse is given by the gradient of the constraint equation: $$\nabla g(x, y) = (2x+y, x+4y)$$We know the gradient is perpendicular to the level curve and points in the direction where the objective function is increasing. Observing that the gradient at the critical points points away from the origin, we conclude that the minimum distance is \(\frac{\sqrt{2}}{\sqrt{5}}\). Since the ellipse is a closed curve and the objective function is a square of the distance (always non-negative), there is no maximum distance. #Answer# The minimum distance between the origin and the ellipse is \(\frac{\sqrt{2}}{\sqrt{5}}\), and there is no maximum distance.

Key Concepts

Ellipse

An ellipse is a fascinating geometric shape, often seen as a stretched circle. It possesses two axes; the longer one is called the major axis, and the shorter one is the minor axis. The standard equation of an ellipse can vary, but in this exercise, we look at a more generalized form:

• The equation \(x^2+xy+2y^2=1\) describes an ellipse not aligned to the main axes.
• It's crucial to understand the ellipse's symmetry and shape to apply mathematical techniques correctly.

Understanding this equation helps us in applying the method of Lagrange multipliers, particularly when dealing with geometrical constraints.

Critical Points

Critical points are essential when trying to find extreme values, like minimum or maximum distances. They are points where the derivative (or gradient) of a function is zero, which indicates a potential minimum or maximum.

• To find them, compute the derivatives of the function and set them to zero.
• In this exercise, we derive the critical points from our Lagrange function \( L(x, y, \lambda) \), leading to equations that help determine where extreme values occur.

Recognizing these points on the ellipse tells us where the distance to the origin can be the shortest or longest. Solving these derivatives provides key insight into the problem.

Gradient

The gradient of a function in multivariable calculus is a vector indicating the direction of steepest ascent. For our case, it helps us understand how the function behaves around specific points.

• The gradient of the constraint \( g(x,y) = x^2 + xy + 2y^2 - 1 \) is \( abla g(x, y) = (2x+y, x+4y) \).
• This vector function is crucial in the Lagrange multiplier method, ensuring the direction of the greatest increase aligns with the distance's level curves.

It helps determine if we have located a minimum (or potential maximum) and ensures that the movement away from this point aligns with the described constraint.

Distance Calculation

Calculating distances in this context involves understanding how far the origin is from the point on the ellipse. We focus on the squared distance to minimize calculation complexity:

• The formula used is \( D^2 = x^2 + y^2 \), representing the square of the distance to the origin, eliminating square roots initially.
• By evaluating this expression at critical points, we find how close these points bring the ellipse to the origin.

Distance calculation informs us of the extreme geographical boundaries (minimum and maximum distances) without direct point-by-point evaluation, making the method efficient and effective.