Optimizing the Dimensions of an Open-Top Rectangular Box to Minimize Surface Area

In many practical scenarios, such as manufacturing and design, it is crucial to find the most efficient dimensions for a container to minimize material usage. This article explores how to optimize the dimensions of an open-top rectangular box that must hold a volume of 108 cubic meters to minimize its surface area.

Step-by-Step Solution

Defining Variables and Constraints

Let's denote the dimensions of the box as follows:

x length of the base y width of the base h height of the box

The volume of the box is given by:

V x cdot y cdot h 108 text{m}^3

From this constraint, we can express the height (h) in terms of the base dimensions (x) and (y)

h frac{108}{x cdot y}

Surface Area Formula

The surface area (S) of an open-top box is given by:

S x cdot y 2xh 2yh

Substituting (h) from the volume constraint into the surface area equation, we get:

S x cdot y 2xleft(frac{108}{x cdot y}right) 2yleft(frac{108}{x cdot y}right)

After simplification, this becomes:

S xy frac{216}{y} frac{216}{x}

Finding Critical Points

To minimize the surface area, we need to find the critical points. We take the partial derivatives of (S) with respect to (x) and (y) and set them to zero.

Partial Derivative w.r.t. (x)

(frac{partial S}{partial x} y - frac{216}{x^2} 0)

(Rightarrow y frac{216}{x^2})

Partial Derivative w.r.t. (y)

(frac{partial S}{partial y} x - frac{216}{y^2} 0)

(Rightarrow x frac{216}{y^2})

Solving the System of Equations

Substituting (y frac{216}{x^2}) into (x frac{216}{y^2}), we get:

x frac{216}{left(frac{216}{x^2}right)^2} frac{216x^4}{216^2} frac{x^4}{216})

(Rightarrow x^3 216)

(Rightarrow x 6 text{ meters})

Substituting (x 6 text{ meters}) back to find (y):

y frac{216}{6^2} frac{216}{36} 6 text{ meters})

Finally, we find the height (h):

h frac{108}{x cdot y} frac{108}{6 cdot 6} frac{108}{36} 3 text{ meters})

Conclusion

The dimensions of the box that minimize the surface area are:

Length: x 6 text{ meters} Width: y 6 text{ meters} Height: h 3 text{ meters}

Therefore, the dimensions of the optimal box are 6 text{ meters} times 6 text{ meters} times 3 text{ meters}.

Mathematical Optimization Using Lagrange Multipliers

Alternatively, we can use the Lagrange multiplier method to solve the problem. We aim to minimize the surface area subject to the volume constraint. The equations are derived from:

( abla S lambda abla V) (V 0)

This results in the following equations:

(S_x lambda V_x)

(S_y lambda V_y)

(S_z lambda V_z)

(V lbh - 108 0)

After substitution and simplification, we obtain:

(l b 6 text{ meters})

(h 3 text{ meters})

Thus, the optimal dimensions of the box are once again confirmed as 6 text{ meters} times 6 text{ meters} times 3 text{ meters}.