Problem 33
Question
Find the point in space the sum of whose coordinates is 48 and whose distance from the origin is minimum.
Step-by-Step Solution
Verified Answer
The point is (24, 24, 0) with distance 24√2.
1Step 1: Understand the Problem
We need to find a point \((x, y, z)\) in 3-dimensional space such that the sum of its coordinates \(x + y + z = 48\), and its distance from the origin \(O = (0, 0, 0)\) is minimized.
2Step 2: Set up the Distance Formula
The distance from any point \((x, y, z)\) to the origin \((0, 0, 0)\) is given by the formula \(d = \sqrt{x^2 + y^2 + z^2}\). Our goal is to minimize this distance given the constraint \(x + y + z = 48\).
3Step 3: Use Constraint to Reduce Dimensions
Since \(x + y + z = 48\), we can express one variable in terms of the others, say \(z = 48 - x - y\). Substitute \(z\) into the distance formula to get\[d = \sqrt{x^2 + y^2 + (48 - x - y)^2}\].
4Step 4: Simplify the Distance Function
Simplify the expression under the square root:\[d = \sqrt{2x^2 + 2y^2 - 96x - 96y + 2304} \].
5Step 5: Find Critical Points
To minimize the function, take the partial derivatives with respect to \(x\) and \(y\) and set them to zero:- \(\frac{\partial d}{\partial x} = \frac{4x - 96}{\sqrt{2x^2 + 2y^2 - 96x - 96y + 2304}} = 0\)- \(\frac{\partial d}{\partial y} = \frac{4y - 96}{\sqrt{2x^2 + 2y^2 - 96x - 96y + 2304}} = 0\)Solve these equations to find \(x = 24\) and \(y = 24\).
6Step 6: Calculate the Third Coordinate
Using the constraint \(x + y + z = 48\), substitute \(x = 24\) and \(y = 24\) to find \(z = 48 - 24 - 24 = 0\).
7Step 7: Verify the Results
Substitute the values back into the distance formula:\[d = \sqrt{24^2 + 24^2 + 0^2} = \sqrt{1152} = \sqrt{48 \times 24} = 24\sqrt{2}\].This is indeed the minimum possible distance given the constraint.
Key Concepts
3-dimensional spacedistance formulapartial derivativesconstraint
3-dimensional space
In this problem, we are exploring 3-dimensional space. This means we are considering a space that has three different dimensions, often represented by the axes labeled as
Think of it like finding a specific location in a room where you need to know: how far along the width, how far up from the floor, and how far into the room the point is.
In our specific exercise, we want to find such a point where the sum of the coordinates equals 48, which adds an extra layer of complexity.
- x (horizontal direction)
- y (vertical direction)
- z (depth/height direction)
Think of it like finding a specific location in a room where you need to know: how far along the width, how far up from the floor, and how far into the room the point is.
In our specific exercise, we want to find such a point where the sum of the coordinates equals 48, which adds an extra layer of complexity.
distance formula
Understanding the distance formula is critical here. The distance from a point \((x, y, z)\) to the origin \((0, 0, 0)\) can be calculated using the formula:
Imagine a diagonal in a cube, where you need to calculate the length of that diagonal. It's similar to finding the longest stick that can fit perfectly across a room when touching two opposite corners at once.
Our task is to minimize this distance, given a specific constraint related to the sum of the coordinates.
- \(d = \sqrt{x^2 + y^2 + z^2}\)
Imagine a diagonal in a cube, where you need to calculate the length of that diagonal. It's similar to finding the longest stick that can fit perfectly across a room when touching two opposite corners at once.
Our task is to minimize this distance, given a specific constraint related to the sum of the coordinates.
partial derivatives
Partial derivatives come into play when you deal with functions of more than one variable, like \(d = \sqrt{2x^2 + 2y^2 - 96x - 96y + 2304}\).
In optimization, we look for critical points by setting these derivatives to zero.
For instance, when we partially differentiate with respect to x and y in our problem, we set:
This is akin to finding peaks and valleys on a landscape by feeling the slopes around you until they flatten out.
In optimization, we look for critical points by setting these derivatives to zero.
For instance, when we partially differentiate with respect to x and y in our problem, we set:
- \(\frac{\partial d}{\partial x} = \frac{4x - 96}{\sqrt{2x^2 + 2y^2 - 96x - 96y + 2304}} = 0\)
- \(\frac{\partial d}{\partial y} = \frac{4y - 96}{\sqrt{2x^2 + 2y^2 - 96x - 96y + 2304}} = 0\)
This is akin to finding peaks and valleys on a landscape by feeling the slopes around you until they flatten out.
constraint
A constraint is a rule or condition that the solution must satisfy. In this exercise, our constraint is that the sum of the coordinates equals 48: \(x + y + z = 48\).
Constraints can simplify complex problems by reducing the number of variables involved.
In this case, we used the constraint to express one variable, z, in terms of the others:
Constraints help narrow down the infinite possibilities in a space, leading to a clear solution by limiting the scope.
Constraints can simplify complex problems by reducing the number of variables involved.
In this case, we used the constraint to express one variable, z, in terms of the others:
- \(z = 48 - x - y\)
Constraints help narrow down the infinite possibilities in a space, leading to a clear solution by limiting the scope.
Other exercises in this chapter
Problem 32
Find an equation of the plane tangent to the graph of the given function at the indicated point(s). $$ f(x, y)=\frac{x+2}{y+1} ;(2,3,1) $$
View solution Problem 32
Find \(f_{x y}\) and \(f_{y x}\) $$ f(x, y, z)=x^{4}-2 x^{2} y \sqrt{z}+3 y z^{4}+2 $$
View solution Problem 33
$$ \text { Let } z=f(x-y) . \text { Show that } \frac{\partial z}{\partial x}=-\frac{\partial z}{\partial y} \text { . } $$
View solution Problem 33
Find an equation of the plane tangent to the graph of the given function at the indicated point(s). $$ g(x, y)=\sin \pi x y ;(-\sqrt{2}, \sqrt{2}, 0) \text { an
View solution