Question

Absolute extrema on open and/or unbounded regions. Find the points on the cone \(z^{2}=x^{2}+y^{2}\) nearest the point \(P(6,8,0)\)

Short answer

Question: Find the nearest points on the cone \(z^2 = x^2 + y^2\) to the point \(P(6, 8, 0)\). Answer: The nearest points on the cone are given by Position 1: \((x, y, z) = \left(\frac{3}{4}y, y, \frac{5}{4}y \right)\) and Position 2: \((x, y, z) = \left(\frac{3}{4}y, y, -\frac{5}{4}y \right)\), where the actual coordinates depend on the value of \(y\).

Step-by-step solution

Compute the distance formula squared

To find the distance squared between any point \((x, y, z)\) on the cone and the point \(P(6, 8, 0)\), we use the distance formula squared: $$D^2 = (x-6)^2 + (y-8)^2 + (z-0)^2$$

Set up the Lagrange multiplier equation

To find the points on the cone that minimize the distance squared, we use the method of Lagrange multipliers. We set up the Lagrange multiplier equation by considering the constraint \(z^2 - x^2 - y^2 = 0\): $$L(x, y, z, \lambda) = (x-6)^2 + (y-8)^2 + z^2 - \lambda(z^2 - x^2 - y^2)$$

Compute the partial derivatives

Compute the partial derivatives of \(L(x, y, z, \lambda)\) with respect to \(x, y, z\), and \(\lambda\): $$\frac{\partial L}{\partial x} = 2(x-6) + 2\lambda x$$ $$\frac{\partial L}{\partial y} = 2(y-8) + 2\lambda y$$ $$\frac{\partial L}{\partial z} = 2z - 2\lambda z$$ $$\frac{\partial L}{\partial \lambda} = z^2 - x^2 - y^2$$

Solve the system of equations

Equate the partial derivatives to zero and solve for \(x, y, z\), and \(\lambda\): $$2(x-6) + 2\lambda x = 0$$ $$2(y-8) + 2\lambda y = 0$$ $$2z - 2\lambda z = 0$$ $$z^2 - x^2 - y^2 = 0$$ At this point, you can either use substitution or a linear algebra method such as row reduction to solve the system. We'll continue this problem with substitution. To do this, we first express \(\lambda\) in terms of \(x\) and \(y\): $$\lambda x = x-6 \Rightarrow \lambda = \frac{x-6}{x}$$ $$\lambda y = y-8 \Rightarrow \lambda = \frac{y-8}{y}$$ Now, equate the two expressions for \(\lambda\) and solve for \(x\): $$\frac{x-6}{x} = \frac{y-8}{y}$$ Since \(x\) is multiplied by the constraint equation, it cannot be equal to 0. Therefore, we can cross-multiply: $$(x-6)y = (y-8)x$$ $$xy - 6y = xy - 8x \Rightarrow x = \frac{3}{4}y$$

Find the value of \(z\)

Substitute the relationship \(x = \frac{3}{4}y\) into the constraint equation and solve for \(z\): $$z^2 - \left(\frac{3}{4}y\right)^2 - y^2 = 0$$ $$z^2 = \frac{9}{16}y^2 + y^2$$ $$z^2 = \frac{25}{16}y^2 \Rightarrow z = \pm \frac{5}{4}y$$

Find the nearest points on the cone

Use the found relationships between \(x, y\), and \(z\) to find the coordinates of the nearest points on the cone: Position 1: \((x, y, z) = \left(\frac{3}{4}y, y, \frac{5}{4}y \right)\) Position 2: \((x, y, z) = \left(\frac{3}{4}y, y, -\frac{5}{4}y \right)\) The actual coordinates would depend on the value of \(y\). However, we have expressed the nearest points on the cone in terms of \(y\), which is the expected result.

Key concepts

Distance Formula

When finding the closest point on a geometric shape to a specific location, the Distance Formula is essential. It helps measure how far apart two points are in space. In the given exercise, we want to find the point on the cone described by the equation \(z^2 = x^2 + y^2\) that is nearest to the point \(P(6, 8, 0)\). To simplify calculations, instead of directly computing the distance, we find the square of the distance.

• The distance squared, often written as \(D^2\), is calculated as: \[(x-6)^2 + (y-8)^2 + (z-0)^2\].
• This equation lets us avoid handling square roots and just focus on easier-to-manipulate algebraic expressions.

Without the square root in the equation, the functions become polynomial, making derivatives easier to manage when applying other methods, like Lagrange Multipliers.

Optimality Conditions

In mathematics, finding the optimal points on a surface or curve under specific constraints is a common problem. The use of Optimality Conditions helps identify when you have reached the best possible outcome, such as the minimum or maximum point.

• To achieve this, we use a method called Lagrange Multipliers, which adds a new variable \(\lambda\) to account for constraints like the one in our exercise: \(z^2 - x^2 - y^2 = 0\).
• The Lagrangian function is then formulated as: \[L(x, y, z, \lambda) = (x-6)^2 + (y-8)^2 + z^2 - \lambda(z^2 - x^2 - y^2)\].
• Once we have the Lagrangian, we take its partial derivatives and set them to zero to find stationary points, which guide us to the optimal solutions.

This approach helps in handling complex constraints while still searching for extreme values like the nearest point on the cone.

Partial Derivatives

To solve optimization problems involving functions of several variables, you often need to use Partial Derivatives. These derivatives reflect how a function changes as one of the variables changes, while the others are held constant. In this exercise, partial derivatives are used in the context of the Lagrangian function.

• Compute partial derivatives with respect to each variable \(x, y, z,\) and the multiplier \(\lambda:\)
• \(\frac{\partial L}{\partial x} = 2(x-6) + 2\lambda x\)
• \(\frac{\partial L}{\partial y} = 2(y-8) + 2\lambda y\)
• \(\frac{\partial L}{\partial z} = 2z - 2\lambda z\)
• \(\frac{\partial L}{\partial \lambda} = z^2 - x^2 - y^2\)

By equating these partial derivatives to zero, we create a system of equations. Solving this system will help us find the values of \(x, y, z,\) and \(\lambda\), which ultimately point to the optimal solution. This method allows us to handle multi-variable functions systematically, ensuring that the constraints are satisfied.