A large container in the shape of a rectangular solid must have a volume of 480 m3. The bottom of the container costs $5/m2 to construct whereas the top and sides cost $3/ m2 to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has the minimum cost

The dimension of the container of this size that has the minimum cost is [tex]x = \sqrt[3]{360}[/tex], [tex]y = \sqrt[3]{360}[/tex] and [tex]z =\frac{4}{3} \sqrt[3]{360}[/tex]

The given parameters are:

Assume the dimensions of the container are: x, y and z.

The volume (V) would be:

[tex]V = xyz[/tex]

Substitute 480 for V

[tex]xyz =480[/tex]

And the objective cost function would be:

[tex]C =8xy + 6xz + 6yz[/tex]

Differentiate the cost function using Lagrange multipliers

[tex]8x + 6z = \lambda xz[/tex]

[tex]8y + 6z = \lambda yz[/tex]

[tex]6x + 6y = \lambda xy[/tex]

Divide equation (1) by (2)

[tex]\frac{8x + 6z}{8y + 6z} = \frac{\lambda xz}{\lambda yz}[/tex]

[tex]\frac{8x + 6z}{8y + 6z} = \frac{x}{y}[/tex]

Factor out 2

[tex]\frac{4x + 3z}{4y + 3z} = \frac{x}{y}[/tex]

Cross multiply

[tex]4xy + 3yz = 4xy + 3xz[/tex]

Evaluate the like terms

[tex]3yz = 3xz[/tex]

Divide both sides by 3z

[tex]y = x[/tex]

Divide the first equation by the third

[tex]\frac{8y + 6z}{6x + 6y} = \frac{\lambda yz}{\lambda xy}[/tex]

[tex]\frac{8y + 6z}{6x + 6y} = \frac{z}{x}[/tex]

Factor out 2

[tex]\frac{4y + 3z}{3x + 3y} = \frac{z}{x}[/tex]

Cross multiply

[tex]4xy+3xz = 3xz + 3yz[/tex]

Cancel out the common terms

[tex]4xy = 3yz[/tex]

Divide both sides by y

[tex]4x = 3z[/tex]

Make z the subject

[tex]z =\frac{4}{3}x\\[/tex]

So, we have:

[tex]y = x[/tex] and [tex]z =\frac{4}{3}x\\[/tex]

Recall that:

[tex]xyz =480[/tex]

Substitute [tex]y = x[/tex] and [tex]z =\frac{4}{3}x\\[/tex]

[tex]x \times x \times \frac 43x = 480[/tex]

So, we have:

[tex]\frac 43x^3 = 480[/tex]

Multiply both sides by 3/4

[tex]x^3 = 360[/tex]

Take the cube roots of both sides

[tex]x = \sqrt[3]{360}[/tex]

Recall that:

[tex]y = x[/tex]

So, we have:

[tex]y = \sqrt[3]{360}[/tex]

Also, we have:

[tex]z =\frac{4}{3}x\\[/tex]

So, we have:

[tex]z =\frac{4}{3} \sqrt[3]{360}[/tex]

Hence, the dimension of the container of this size that has the minimum cost is [tex]x = \sqrt[3]{360}[/tex], [tex]y = \sqrt[3]{360}[/tex] and [tex]z =\frac{4}{3} \sqrt[3]{360}[/tex]

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