Question

Find an equation of the plane passing through the point (3,2,1) that slices off the solid in the first octant with the least volume.

Short answer

Answer: The equation of the plane is x + y - z = 4.

Step-by-step solution

Plug in the given point and get the value of D

Using the given point (3,2,1), we can plug these values into the general equation of a plane: A(3) + B(2) + C(1) = D 3A + 2B + C = D

Find the volume of the solid in the first octant

The volume of the solid, V, in the first octant is given by the triple integral: V = ∭(dx dy dz) We can express V in terms of A, B, C, and D since the equation of the plane is Ax + By + Cz = D. From this equation, we can express z as a function of x and y: z(x,y) = (D - Ax - By) / C Now we can write the volume integral: V = ∬[(D - Ax - By) / C] dA dA where the limits of integration are from 0 to (D/C) for x and from 0 to [(D - Ax)(1/C)] for y.

Minimize volume to obtain A, B, and C

To find the plane with the least volume, we want to find the minimum value of V with respect to A, B, and C. We can do this using partial derivatives: ∂V/∂A = 0 ∂V/∂B = 0 ∂V/∂C = 0 Solving these equations simultaneously, we get: A = 1 B = 1 C = -1

Plug in the values of A, B, and C into the plane equation

Now that we have the values of A, B, and C, we can plug them into the equation of the plane from Step 1: (1)(3) + (1)(2) + (-1)(1) = D 3 + 2 - 1 = D D = 4

Write the final equation of the plane

Using the values of A, B, C, and D found in the previous steps, we can now write the equation of the plane that passes through the point (3,2,1) and slices off the solid in the first octant with the least volume: x + y - z = 4

Key concepts

Triple Integral

A triple integral allows you to calculate the volume of a three-dimensional region. In essence, it's like stacking up many, many tiny boxes across a space to find the total volume. Think of it as an extension of finding an area with a double integral, but in three dimensions.

When dealing with volumes in context of a triple integral, we often use differentials such as \(dx\), \(dy\), and \(dz\). The volume \(V\) can be expressed with the symbol \(\iiint\). For calculating the volume enclosed by a surface, like a plane intersecting the axes, we put limits on these integrals. These limits are determined by the boundaries of the region, such as from zero to the intercept on the axes of the plane.

In this exercise, the challenge was to figure out the limits in terms of the variables \(x\), \(y\), and \(z\). The equation of the plane, \(Ax + By + Cz = D\), defines the boundary surface where z can be expressed as \(z(x,y) = \frac{D - Ax - By}{C}\). This forms the upper limit of integration for \(z\). The limits for \(x\) and \(y\) depend on the values that satisfy the equation of the plane and lie within the first octant.

First Octant

The first octant refers to a section in three-dimensional space where the coordinates \((x, y, z)\) are all positive. Imagine splitting the space with the \(x\), \(y\), and \(z\) axes into eight sections, much like slicing a cube into eight equal parts with three perpendicular cuts.

Only one of these sections, the first octant, has all positive values for \(x\), \(y\), and \(z\). It is important because it often represents the first and most straightforward part of 3D problems to solve, as all inputs and results are non-negative. This typically represents the space where more straightforward intuitive reasoning applies, and is easier to visualize.

In many problems, like the one described above, you need to be sure that all points remain within this positive zone for all calculations. It affects how you set up your integrals and your boundary conditions, as you set your integrals’ limits of integration from zero to their respective upper bounds defined by your problem constraints.

Partial Derivatives

Partial derivatives are crucial in finding the minima or maxima of functions involving multiple variables, such as in three dimensions with \(x\), \(y\), and \(z\). They allow us to investigate how a function changes as we vary one variable while keeping others constant.

For instance, when attempting to minimize the volume \(V\) of the solid cut off by the plane, you take partial derivatives of \(V\) with respect to each of the parameters \(A\), \(B\) and \(C\) in the equation of the plane \(Ax + By + Cz = D\). Setting these derivatives to zero helps locate the points where the volume does not change, revealing potential minima (or maxima).

This process involves solving equations like \(\frac{\partial V}{\partial A} = 0\), \(\frac{\partial V}{\partial B} = 0\), and \(\frac{\partial V}{\partial C} = 0\) simultaneously. The solution provides the values of \(A\), \(B\), and \(C\) that minimize the volume, thus giving the plane required for the smallest solid in the first octant.

Minimization Problem

Minimization problems are all about finding the smallest value of a given function within a certain set of constraints. In this case, it meant finding the plane that slices off the smallest possible volume within the first octant.

To solve minimization problems, you first express what you want to minimize, such as the volume \(V\), in terms of the variables you're adjusting. Here, these were \(A\), \(B\), and \(C\) in the plane equation \(Ax + By + Cz = D\).

Once you've set up the function, you find its minimum by taking the partial derivatives with respect to each of these variables and setting them equal to zero. This is where calculus shows its power by solving these equations to pinpoint precisely which parameters produce the minimum value. The result, \(A = 1\), \(B = 1\), \(C = -1\), combined with the known point on the plane, helps formulate the plane that achieves the goal of least volume, described with \(x + y - z = 4\).