# How do you find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 4?

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To find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane (x + 2y + 3z = 4), we need to maximize the volume function subject to the constraint given by the equation of the plane.

The volume of a rectangular box is given by the product of its three dimensions: length, width, and height.

Let the dimensions of the box be (x), (y), and (z). Since the box is in the first octant, all dimensions must be positive.

We want to maximize the volume, which is (V = xyz), subject to the constraint (x + 2y + 3z = 4).

Using the constraint equation to express (x) in terms of (y) and (z), we have (x = 4 - 2y - 3z).

Substituting this expression for (x) into the volume function, we obtain: [V(y, z) = (4 - 2y - 3z)y(z)]

To maximize (V(y, z)), we take partial derivatives with respect to (y) and (z), set them equal to zero, and solve for (y) and (z).

[\frac{\partial V}{\partial y} = 4z - 6y - 4z^2 = 0] [\frac{\partial V}{\partial z} = 4y - 6z - 4y^2 = 0]

Solving these equations simultaneously gives the values of (y) and (z).

After finding (y) and (z), we can find (x) using the constraint equation (x = 4 - 2y - 3z).

Finally, we calculate the volume (V = xyz) using the values of (x), (y), and (z). This will give us the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane (x + 2y + 3z = 4).

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