Question

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

Answer 1

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

The given parameters are:

• Volume = 480m^3
• Cost: $5 per square meter for the bottom, and $3 per square meter for the sides.

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

The volume (V) would be:

$V = xyz$

Substitute 480 for V

$xyz =480$

And the objective cost function would be:

$C =8xy + 6xz + 6yz$

Differentiate the cost function using Lagrange multipliers

$8x + 6z = \lambda xz$

$8y + 6z = \lambda yz$

$6x + 6y = \lambda xy$

Divide equation (1) by (2)

$\frac{8x + 6z}{8y + 6z} = \frac{\lambda xz}{\lambda yz}$

$\frac{8x + 6z}{8y + 6z} = \frac{x}{y}$

Factor out 2

$\frac{4x + 3z}{4y + 3z} = \frac{x}{y}$

Cross multiply

$4xy + 3yz = 4xy + 3xz$

Evaluate the like terms

$3yz = 3xz$

Divide both sides by 3z

$y = x$

Divide the first equation by the third

$\frac{8y + 6z}{6x + 6y} = \frac{\lambda yz}{\lambda xy}$

$\frac{8y + 6z}{6x + 6y} = \frac{z}{x}$

Factor out 2

$\frac{4y + 3z}{3x + 3y} = \frac{z}{x}$

Cross multiply

$4xy+3xz = 3xz + 3yz$

Cancel out the common terms

$4xy = 3yz$

Divide both sides by y

$4x = 3z$

Make z the subject

$z =\frac{4}{3}x$

So, we have:

$y = x$ and $z =\frac{4}{3}x$

Recall that:

$xyz =480$

Substitute $y = x$ and $z =\frac{4}{3}x$

$x \times x \times \frac 43x = 480$

So, we have:

$\frac 43x^3 = 480$

Multiply both sides by 3/4

$x^3 = 360$

Take the cube roots of both sides

$x = \sqrt[3]{360}$

Recall that:

$y = x$

So, we have:

$y = \sqrt[3]{360}$

Also, we have:

$z =\frac{4}{3}x$

So, we have:

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

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

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