Problem 62 - Calculus Early Transcendentals

Question

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\).

Step-by-Step Solution

Answer

If so, what are the dimensions? Answer: No, there is no maximum volume for the rectangular box in this case.

Step 1: Write the volume function for the rectangular box

The box has one vertex at the origin, so its dimensions are \(x, y,\) and \(z\). Therefore, the volume function for the rectangular box is given by: $$V(x,y,z) = xyz$$

Step 2: Write the constraint (ellipsoid equation)

The constraint is given by the ellipsoid equation that the vertex on the ellipsoid must satisfy: $$36 x^{2}+4 y^{2}+9 z^{2}=36$$ We can divide both sides of the equation by 36 to get $$x^{2}+\frac{y^{2}}{9}+\frac{z^{2}}{4}=1$$

Step 3: Use Lagrange multipliers to solve for the maximum volume dimensions

We will now use the method of Lagrange multipliers to solve for the maximum volume dimensions. Let \(\lambda\) be the Lagrange multiplier. The function to be maximized is $$L(x,y,z,\lambda)=xyz+\lambda \left(x^{2}+\frac{y^{2}}{9}+\frac{z^{2}}{4}-1\right)$$ Now find the partial derivatives: \begin{align*} \frac{\partial L}{\partial x} &= yz + 2\lambda x \\ \frac{\partial L}{\partial y} &= xz + \frac{2}{9}\lambda y \\ \frac{\partial L}{\partial z} &= xy + \frac{1}{2}\lambda z \\ \frac{\partial L}{\partial \lambda} &= x^{2}+\frac{y^{2}}{9}+\frac{z^{2}}{4}-1 \end{align*} We need to solve for \(x,y,z,\) and \(\lambda\) by setting these four partial derivatives equal to zero.

Step 4: Solve the system of equations for \(x, y, z,\) and \(\lambda\)

From the partial derivatives, we have the 4 equations: \begin{align*} yz + 2\lambda x &= 0 \\ xz + \frac{2}{9}\lambda y &= 0 \\ xy + \frac{1}{2}\lambda z &= 0 \\ x^{2}+\frac{y^{2}}{9}+\frac{z^{2}}{4}-1 &= 0 \end{align*} Multiplying (1) and (3) yields: $$xyz+2xyz\lambda^2=0$$ So \(xyz(1+2\lambda^2)=0\). Since the volume cannot be zero, it implies that \(\lambda^2=-\frac{1}{2}\). However, since \(\lambda^2 \geq 0\), this contradicts the condition of the Lagrange multiplier method. Therefore, there is no maximum volume in this case. The rectangular box in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\) has no maximum volume.