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Chapter 14: Problem 49

Among all the points on the graph of \(z=10-x^{2}-y^{2}\) that lie above the plane \(x+2 y+3 z=0\) , find the point farthest from the plane.

Short Answer

Expert verified
The point farthest from the plane is \(\left(\frac{1}{6}, \frac{1}{3}, \frac{347}{36}\right)\).

Step by step solution

01

Set Up the Problem

We are given a function \(z=10-x^2-y^2\) and a plane \(x+2y+3z=0\). We need to find the point on the surface of the graph of the function that is farthest from the plane.
02

Understand Distance to the Plane

The distance \(D\) from a point \((x, y, z)\) to a plane \(ax+by+cz+d=0\) is given by the formula: \[ D = \frac{|ax+by+cz+d|}{\sqrt{a^2+b^2+c^2}}. \] For the plane \(x+2y+3z=0\), the distance becomes \[ D = \frac{|x + 2y + 3z|}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{|x + 2y + 3z|}{\sqrt{14}}. \]
03

Formulate the Function to Maximize

To maximize the distance, substitute \(z=10-x^2-y^2\) into the distance formula: \[ D = \frac{|x + 2y + 3(10-x^2-y^2)|}{\sqrt{14}} = \frac{|x + 2y + 30-3x^2-3y^2|}{\sqrt{14}}. \] To simplify, consider maximizing the expression \(x + 2y + 30 - 3x^2 - 3y^2\).
04

Use Calculus to Find Critical Points

To find the critical points, take partial derivatives with respect to \(x\) and \(y\), set them to zero, and solve: - \( \frac{\partial}{\partial x} = 1 - 6x = 0 \) yields \( x = \frac{1}{6} \).- \( \frac{\partial}{\partial y} = 2 - 6y = 0 \) yields \( y = \frac{1}{3} \).
05

Find the Corresponding z-value

Substitute \(x=\frac{1}{6}\) and \(y=\frac{1}{3}\) into \(z=10-x^2-y^2\) to find the \(z\)-value: \[ z = 10 - \left(\frac{1}{6}\right)^2 - \left(\frac{1}{3}\right)^2 = 10 - \frac{1}{36} - \frac{1}{9} = \frac{347}{36}. \]
06

Verify with Second Derivative Test

Use second derivative test or analyze the behavior to ensure the point found gives a maximum value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distance to a Plane
When dealing with three-dimensional geometry, understanding the distance from a point to a plane is crucial. The distance formula starts with a plane equation given by:
  • If your plane is represented as \( ax + by + cz + d = 0 \), and the point identified as \( (x, y, z) \), the formula for distance \( D \) is:\[ D = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}} \]
This formula results from projecting the vector from a point to a plane onto the perpendicular.
For example, if you have a specific plane \( x + 2y + 3z = 0 \), it simplifies to:
\[ D = \frac{|x + 2y + 3z|}{\sqrt{14}}. \]
  • The numerator measures direct projection, and the denominator adjusts for plane's orientation.
Partial Derivatives
Partial derivatives are used to find how a function changes as one of its input variables change, while keeping the other variables constant.
These are essential in optimization problems when dealing with a function of several variables. For instance:
  • In our case, take the function \( z = 10 - x^2 - y^2 \).
  • To find how this changes with \( x \), compute \( \frac{\partial}{\partial x} = -2x \). Similarly with \( y \), it's \( \frac{\partial}{\partial y} = -2y \).
Partial derivatives are like looking at one particular slice of a larger, multi-dimensional problem.
Critical Points
Finding the critical points of a function helps determine where it may achieve its maximum or minimum values.
  • To find these points, we set the gradients (or partial derivatives) to zero.
  • In our problem, derivatives are set as:\( \frac{\partial}{\partial x} = 1 - 6x \) and \( \frac{\partial}{\partial y} = 2 - 6y \).
  • Solving these gives us potential extrema at \( x = \frac{1}{6} \) and \( y = \frac{1}{3} \).
These points act like signposts, navigating us to where functions behave interestingly.
Second Derivative Test
This test helps identify if a critical point is a maximum, minimum, or neither by examining the curvature at that point.
  • First, compute the second partial derivatives of the function:
  • \( \frac{\partial^2}{\partial x^2} \) and \( \frac{\partial^2}{\partial y^2} \). In quadratic functions, this indicates concavity.
  • In practice, these values help discern local maxima or minima by confirming if critical points sit in peaks or troughs.
Understanding this test makes identifying function behavior more visual and intuitive, reducing critical point ambiguity.

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Most popular questions from this chapter

In Exercises \(25-30,\) find the linearization \(L(x, y)\) of the function at each point. $$ f(x, y)=e^{x} \cos y \text { at } \quad \text { a. }(0,0), \quad \text { b. }(0, \pi / 2) $$

Change along a helix Find the derivative of \(f(x, y, z)=\) \(x^{2}+y^{2}+z^{2}\) in the direction of the unit tangent vector of the helix $$\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+t \mathbf{k}$$ at the points where \(t=-\pi / 4,0,\) and \(\pi / 4 .\) The function \(f\) gives the square of the distance from a point \(P(x, y, z)\) on the helix to the origin. The derivatives calculated here give the rates at which the square of the distance is changing with respect to \(t\) as \(P\) moves through the points where \(t=-\pi / 4,0,\) and \(\pi / 4\) .

In Exercises \(31-36,\) find the linearization \(L(x, y)\) of the function \(f(x, y)\) at \(P_{0} .\) Then find an upper bound for the magnitude \(|E|\) of the error in the approximation \(f(x, y) \approx L(x, y)\) over the rectangle \(R\) . $$ \begin{array}{l}{f(x, y)=x^{2}-3 x y+5 \text { at } P_{0}(2,1)} \\ {R :|x-2| \leq 0.1, \quad|y-1| \leq 0.1}\end{array} $$

Each of Exercises \(63-66\) gives a function \(f(x, y, z)\) and a positive number \(\epsilon .\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y, z)\) , $$ \quad \sqrt{x^{2}+y^{2}+z^{2}}<\delta \Rightarrow|f(x, y, z)-f(0,0,0)|<\epsilon $$ $$ f(x, y, z)=x^{2}+y^{2}+z^{2}, \quad \epsilon=0.015 $$

The Sandwich Theorem for functions of two variables states that if \(g(x, y) \leq f(x, y) \leq h(x, y)\) for all \((x, y) \neq\left(x_{0}, y_{0}\right)\) in a disk centered at \(\left(x_{0}, y_{0}\right)\) and if \(g\) and \(h\) have the same finite limit \(L\) as \((x, y) \rightarrow\left(x_{0}, y_{0}\right),\) then $$ \lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L $$ Use this result to support your answers to the questions in Exercises \(45-48 .\) Does knowing that $$ 2|x y|-\frac{x^{2} y^{2}}{6}<4-4 \cos \sqrt{|x y|}<2|x y| $$ tell you anything about $$ \lim _{(x, y) \rightarrow(0,0)} \frac{4-4 \cos \sqrt{|x y|}}{|x y|} ? $$ Give reasons for your answer.
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