Question

Use Lagrange multipliers to give an alternate solution the indicated exercise in Section\(\;{\bf{11}}.{\bf{7}}\)Exercise\({\bf{42}}\).

Short answer

Largest volume is\(V = \frac{{{L^3}}}{{3\sqrt 3 }}.\)

Step-by-step solution

Method of Lagrange multipliers

To find the maximum and minimum values of\(f(x,y,z)\)subject to the constraint\(g(x,y,z) = k\)(assuming that these extreme values exist and\(\nabla g \ne {\bf{0}}\)on the surface\(g(x,y,z) = k)\):

(a) Find all values of\(x,y,z\), and\(\lambda \)such that\(\nabla f(x,y,z) = \lambda \nabla g(x,y,z)g(x,y,z) = k\)

And

(b) Evaluate\(f\)at all the points\((x,y,z)\)that result from step (a). The largest of these values is the maximum value of\(f\); the smallest is the minimum value of\(f\).

Assumption

Let \(V\)be the volume and define the dimensions as\(x,y\), and\(z\). So, the volume of the box is\(V = xyz\). Let\(D\) be the length of longest diagonal.Apply the distance formula to get\(D = \sqrt {{x^2} + {y^2} + {z^2}} \).

Let \(D = L\)\( \Rightarrow {D^2} = {L^2}\)

Apply Lagrange multiplier

The method of Lagrange multipliers for constrained optimization is to solve the system of equations given by\(\nabla f = \lambda \nabla g\,\,,g = k\)

Here\(f = xyz\,\,\& \,\,g = {x^2} + {y^2} + {z^2}\)

Substituting & comparing we get:

\(\begin{array}{l}yz = 2\lambda x \ldots \ldots (1)\\xz = 2\lambda y \ldots \ldots (2)\\xy = 2\lambda z \ldots \ldots (3)\\{x^2} + {y^2} + {z^2} = {L^2} \ldots (4)\end{array}\)

Solve the system

Solve equations (1), (2), and (3) for\(\lambda \).

\(\begin{array}{l}2\lambda = \frac{{yz}}{x} \ldots \ldots (5)\\2\lambda = \frac{{xz}}{y} \ldots \ldots (6)\\2\lambda = \frac{{xy}}{z} \ldots \ldots (7)\end{array}\)

Equate equations (5) and (6) and solve.

\(\begin{array}{c}\frac{{yz}}{x} = \frac{{xz}}{y}\\{y^2}z = {x^2}z\\x = y\end{array}\)

Equate equations (5) and (7) and solve.

\(\begin{array}{c}\frac{{yz}}{x} = \frac{{xy}}{z}\\y{z^2} = {x^2}y\\x = z\end{array}\)

Using equation (4) and solve for\(x\).

\(\begin{array}{l}3{x^2} = {L^2}\\x = \frac{L}{{\sqrt 3 }}{\rm{ Since }}x = y{\rm{ and }}x = z\end{array}\)

Compute volume

Compute and simplify the volume of the largest possible box:

\(\begin{array}{c}V = xyz = {x^3}{\rm{ Since }}x = y = z\\ = \frac{{{L^3}}}{{3\sqrt 3 }}\end{array}\)

Hence, the largest possible volume is\(V = \frac{{{L^3}}}{{3\sqrt 3 }}.\)