Math Problem Statement

Find the minimal distance from the point (1, 4, 0) to the surface z^2 = 2x y + y^2.

Solution

To find the minimal distance from the point (1,4,0)(1, 4, 0) to the surface defined by the equation z2=2xy+y2z^2 = 2xy + y^2, we can use the method of Lagrange multipliers. This involves minimizing the distance function subject to the constraint provided by the surface equation.

Step 1: Distance Function

The distance DD from any point (x,y,z)(x, y, z) on the surface to the point (1,4,0)(1, 4, 0) is given by:

D(x,y,z)=(x1)2+(y4)2+z2D(x, y, z) = \sqrt{(x - 1)^2 + (y - 4)^2 + z^2}

However, to simplify the optimization, we can minimize the square of the distance function:

f(x,y,z)=(x1)2+(y4)2+z2f(x, y, z) = (x - 1)^2 + (y - 4)^2 + z^2

Step 2: Constraint

The surface is given by the equation:

g(x,y,z)=z22xyy2=0g(x, y, z) = z^2 - 2xy - y^2 = 0

Step 3: Lagrange Multipliers

To minimize f(x,y,z)f(x, y, z) subject to the constraint g(x,y,z)=0g(x, y, z) = 0, we introduce a Lagrange multiplier λ\lambda and solve the system:

f(x,y,z)=λg(x,y,z)\nabla f(x, y, z) = \lambda \nabla g(x, y, z)

This gives us the following equations:

  1. fx=2(x1)=λgx=λ(2y)\frac{\partial f}{\partial x} = 2(x - 1) = \lambda \frac{\partial g}{\partial x} = \lambda(-2y)
  2. fy=2(y4)=λgy=λ(2x2y)\frac{\partial f}{\partial y} = 2(y - 4) = \lambda \frac{\partial g}{\partial y} = \lambda(-2x - 2y)
  3. fz=2z=λgz=λ(2z)\frac{\partial f}{\partial z} = 2z = \lambda \frac{\partial g}{\partial z} = \lambda(2z)
  4. g(x,y,z)=z22xyy2=0g(x, y, z) = z^2 - 2xy - y^2 = 0

Step 4: Solving the System

From equation (3):

2z=λ(2z)2z = \lambda(2z)

If z0z \neq 0, then λ=1\lambda = 1. If z=0z = 0, then the constraint reduces to 0=2xyy20 = -2xy - y^2, which simplifies to y(2x+y)=0y(2x + y) = 0.

Let's first consider z=0z = 0:

  • If y=0y = 0, then x=1x = 1, but this does not satisfy the constraint 2xy+y2=02xy + y^2 = 0.
  • If 2x+y=02x + y = 0, then y=2xy = -2x, and substituting this into the constraint, we have 0=00 = 0, so y=2xy = -2x.

Now we can solve equations (1) and (2) under the assumption λ=1\lambda = 1 and check for both cases.

Step 5: Special Cases and Check the Solutions

For z=0z = 0, and y=2xy = -2x:

  • Plugging back into f(x,y,z)=(x1)2+(y4)2+z2f(x, y, z) = (x - 1)^2 + (y - 4)^2 + z^2 to compute the minimum value.

For other cases where z0z \neq 0, use λ=1\lambda = 1 and solve the system.

Finally, the correct minimal distance will be determined by the minimum value of f(x,y,z)f(x, y, z) among the feasible solutions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Lagrange multipliers
Distance formula

Formulas

Distance formula
Lagrange multipliers

Theorems

Lagrange multipliers theorem

Suitable Grade Level

Undergraduate level