Problem 7
Question
Find the largest box that will fit in the positive octant \((x \geq 0, y \geq 0,\) and \(z \geq 0)\) and underneath the plane \(z=12-2 x-3 y\).
Step-by-Step Solution
Verified Answer
The largest box has volumes \(\frac{16}{3}\) fitting in the positive octant under the plane.
1Step 1: Understand the Problem
We need to find the largest box that can fit in the positive octant (where all coordinates are non-negative) and be underneath the plane described by the equation \(z = 12 - 2x - 3y\). The volume of the box is what we need to maximize.
2Step 2: Define the Box
Let the dimensions of the box be \(x\), \(y\), and \(z\). Thus, the volume \(V\) of the box is \(V = x \, y \, z\). According to the problem, the box must be entirely below or on the plane, so \(z = 12 - 2x - 3y\). Substitute this into the volume equation to express \(V\) in terms of \(x\) and \(y\): \[ V = x \, y \, (12 - 2x - 3y) \] This simplifies to \[ V = 12xy - 2x^2y - 3xy^2 \].
3Step 3: Find Critical Points
We need to find the critical points of the volume function \(V(x, y) = 12xy - 2x^2y - 3xy^2\). First, find the partial derivatives: \( \frac{\partial V}{\partial x} = 12y - 4xy - 3y^2 \) and \( \frac{\partial V}{\partial y} = 12x - 2x^2 - 6xy \). Set both partial derivatives to 0 and solve the system of equations to find critical points.
4Step 4: Solve the System of Equations
Solving \( 12y - 4xy - 3y^2 = 0 \) gives \[ y(12 - 4x - 3y) = 0 \]. This implies \(y = 0\) or \(12 - 4x - 3y = 0\).Solving \( 12x - 2x^2 - 6xy = 0 \) gives \[ x(12 - 2x - 6y) = 0 \]. This implies \(x = 0\) or \(12 - 2x - 6y = 0\).For non-zero \(x\) and \(y\), solve simultaneously: \(12 - 4x - 3y = 0\) and \(12 - 2x - 6y = 0\).
5Step 5: Solve the Simultaneous Equations
From \(12 - 4x - 3y = 0\), express \(y\): \[ y = \frac{12 - 4x}{3} \]. Substitute \(y\) in \(12 - 2x - 6y = 0\): \[ 12 - 2x - 6\left(\frac{12 - 4x}{3}\right) = 0 \]. Which simplifies to: \[ 2x = 2 \] \( x = 1 \). Then \( y = \frac{12 - 4(1)}{3} = \frac{8}{3} \).
6Step 6: Calculate the Volume and Compare
With \(x = 1\) and \(y = \frac{8}{3}\), find \(z\) using the plane equation: \( z = 12 - 2(1) - 3\left(\frac{8}{3}\right) = 2 \). The volume is \( V = 1 \times \frac{8}{3} \times 2 = \frac{16}{3} \). Check boundary conditions like \(x = 0\) or \(y = 0\), which result in zero volume, affirming our maximum.
Key Concepts
Volume MaximizationPartial DerivativesCritical Points
Volume Maximization
Volume maximization involves finding the largest possible volume for a given shape under certain constraints. In calculus, this often means using functions to model the situation and applying specific techniques to find the maximum values.
In this exercise, we're tasked with maximizing the volume of a box that can fit under a plane in the positive octant. This geometrical setup can be viewed as a classic application of optimization in calculus. We first define the volume of the box as a function of its dimensions, specifically as \( V = x \, y \, z \). However, due to the constraint given by the plane \( z = 12 - 2x - 3y \), we substitute \( z \) in our volume equation to express it entirely in terms of \( x \) and \( y \).
This yields the function \( V = 12xy - 2x^2y - 3xy^2 \), which describes how the volume changes as the shape of the box changes. Our goal is to adjust \( x \) and \( y \) to maximize \( V \) without breaking the restriction imposed by the plane.
In this exercise, we're tasked with maximizing the volume of a box that can fit under a plane in the positive octant. This geometrical setup can be viewed as a classic application of optimization in calculus. We first define the volume of the box as a function of its dimensions, specifically as \( V = x \, y \, z \). However, due to the constraint given by the plane \( z = 12 - 2x - 3y \), we substitute \( z \) in our volume equation to express it entirely in terms of \( x \) and \( y \).
This yields the function \( V = 12xy - 2x^2y - 3xy^2 \), which describes how the volume changes as the shape of the box changes. Our goal is to adjust \( x \) and \( y \) to maximize \( V \) without breaking the restriction imposed by the plane.
Partial Derivatives
Partial derivatives allow us to study how changing one variable affects a function while keeping other variables constant. They provide a way to explore how a multivariable function behaves and are essential in finding points where maximum or minimum values occur.
For the volume function \( V(x, y) = 12xy - 2x^2y - 3xy^2 \), we need to use partial derivatives to understand the sensitivity of \( V \) with respect to changes in \( x \) and \( y \). By computing \( \frac{\partial V}{\partial x} = 12y - 4xy - 3y^2 \) and \( \frac{\partial V}{\partial y} = 12x - 2x^2 - 6xy \), we find expressions that show how small changes in these dimensions affect the overall volume.
Setting these derivatives to zero helps us identify critical points, indicating where the function could potentially have a maximum or minimum. In simple terms, these derivatives help us navigate the shape of the function's surface to uncover the best possible outcomes under given constraints.
For the volume function \( V(x, y) = 12xy - 2x^2y - 3xy^2 \), we need to use partial derivatives to understand the sensitivity of \( V \) with respect to changes in \( x \) and \( y \). By computing \( \frac{\partial V}{\partial x} = 12y - 4xy - 3y^2 \) and \( \frac{\partial V}{\partial y} = 12x - 2x^2 - 6xy \), we find expressions that show how small changes in these dimensions affect the overall volume.
Setting these derivatives to zero helps us identify critical points, indicating where the function could potentially have a maximum or minimum. In simple terms, these derivatives help us navigate the shape of the function's surface to uncover the best possible outcomes under given constraints.
Critical Points
Critical points are where the partial derivatives of a function are zero or undefined, indicating a potential "turning point" in the behavior of the function. In optimization problems, we examine these points to determine whether they represent maximum or minimum values.
In our exercise, after computing the partial derivatives, we set them to zero to identify pairs \( (x, y) \) that could maximize the volume \( V(x, y) \). Solving these equations, \( 12y - 4xy - 3y^2 = 0 \) and \( 12x - 2x^2 - 6xy = 0 \), leads us to find the conditions under which the box's dimensions result in maximum volume.
We explore these equations under the given constraints, noting that solutions where either variable is zero imply no box exists, as the volume would collapse to zero. Therefore, we specifically focus on non-zero solutions to find that the critical point \( x = 1 \) and \( y = \frac{8}{3} \) yields the largest volume, ensuring the maximum capacity of the box within the defined space under the plane.
In our exercise, after computing the partial derivatives, we set them to zero to identify pairs \( (x, y) \) that could maximize the volume \( V(x, y) \). Solving these equations, \( 12y - 4xy - 3y^2 = 0 \) and \( 12x - 2x^2 - 6xy = 0 \), leads us to find the conditions under which the box's dimensions result in maximum volume.
We explore these equations under the given constraints, noting that solutions where either variable is zero imply no box exists, as the volume would collapse to zero. Therefore, we specifically focus on non-zero solutions to find that the critical point \( x = 1 \) and \( y = \frac{8}{3} \) yields the largest volume, ensuring the maximum capacity of the box within the defined space under the plane.
Other exercises in this chapter
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