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?

Question

Evelyn Agard · Accepted Answer

# ext{Volume}\ \ =\ \ 32/81# # ext{cubic units}#

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That's it!

Scarlett Allison · Answer

undefined 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 $x + 2y + 3z = 4$.