91Ó°ÊÓ

Supposefandgare two functions such that they both have continuous partial derivatives.

Let$\left(x_0,y_0,z_0\right)\mathbf{∈}\mathbf{S}\mathbf{:}\mathbf{=}\left\{\left(x,y,z\right):g\left(x,y,z\right)=0\right\}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∇}\mathbf{g}\left(x_0,y_0,z_0\right)\mathbf{≠}\mathbf{0}$.

If the function fcontains a local minimum or local maximum at a point$\left(x_0,y_0,z_0\right)$then there exists $\mathit{λ}\mathbf{∈}\mathit{R}$such that:

$\mathbf{∇}\mathit{f}\left(x_0,y_0,z_0\right)\mathbf{=}\mathit{λ}\mathbf{∇}\mathit{g}\left(x_0,y_0,z_0\right)$

To determine the extremism points, we generally consider the below equations:

role="math" localid="1659235652953" $\mathbf{∇}\mathit{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\mathit{λ}\mathit{g}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$

$\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{g}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\mathbf{0}$

By solving these equations, we get the values of unknown variables, say$\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{}\mathbf{and}\mathbf{}\mathbf{λ}$.

Thus, we will get the local extremism points through the solutions of the above set of equations.

From the given information, it is a typical optimization problem with constraint, where the vertex of the box is constrained on the plane 2x + 3y + 4z = 6 and they need to find the maximize volume V of the box.

So, will proceed in this problem using the Lagrange multiplier method is given

below:

2x + 3y + 4z = 6

Now invoke the Lagrange multiplier method.

$F=xyz+\mathrm{λ}\left(2x+3y+4z\right)$ …... (1)

Take the partial derivative with respect to , and .

$\frac{∂F}{∂x}=yz+2\mathrm{λ}=0$ …... (2)

$\frac{∂F}{∂x}=yz+3\mathrm{λ}=0$ …... (3)

$\frac{∂F}{∂x}=yz+4\mathrm{λ}=0$ …... (4)

Now with three partial derivative equations and the original constraints equation will be solve for unknown proportions.

Now, with the use of equation (2), $\mathrm{λ}=-\frac{yz}{2}$.

Substitute for $\mathrm{λ}$in equation (3).

$$\begin{gathered} xz+3×-\frac{yz}{2}=0 \\ z\left(x-\frac{3}{2}y\right)=0 \\ x=\frac{3}{2}y \end{gathered}$$

Now, substitute in equation (4).

$$\begin{gathered} y×\left(\frac{3}{2}y\right)+4×\left(\frac{-yz}{2}\right)=0 \\ y\left(\frac{3}{2}y-2z\right)=0 \\ y=\frac{4z}{3} \\ z=\frac{3}{4}y \end{gathered}$$

Finally, substitute in equation (1) for x and z.

$$\begin{gathered} 2×\left(\frac{3y}{2}\right)+3y+4×\frac{3}{4}y=6 \\ y=\frac{2}{3} \\ x=1 \\ z=\frac{1}{2} \end{gathered}$$

Therefore, $V=xyz=1×\frac{2}{3}×\frac{1}{2}=\frac{1}{3}$.

Thus, the maximum value of the box is $V=\frac{1}{3}$.