Maximizing the Volume of a Rectangular Box with a Square Base and No Top

Congratulations, you're reading an article on how to optimize the volume of a rectangular box with a square base and no top using a constraint in the form of a given surface area. This is a classic optimization problem in calculus and geometry, and we'll explore the mathematical steps and insights to find the best solution.

Surface Area and Volume Formulas

Let us define the problem clearly. We want to create a rectangular box with a square base and no top. We are given 4 m2 of material to use for the construction. We need to determine the dimensions of the box that will yield the maximum volume.

Mathematical Approach

Defining Variables

We start by defining the variables:x - Length and width of the square base.h - Height of the box.

Surface Area Constraint

The surface area (S) of the box includes the base area and the areas of the four sides (since there is no top). The total surface area is given by:

[ S x^2 4xh 4 ]

Volume of the Box

The volume (V) of the box is:

[ V x^2h ]

Expressing (h) in Terms of (x)

From the surface area equation, we can solve for (h):

[ 4xh 4 - x^2 implies h frac{4 - x^2}{4x} ]

Substituting (h) into the Volume Equation

Now, substitute (h) into the volume equation:

[ V x^2 cdot frac{4 - x^2}{4x} frac{x(4 - x^2)}{4} frac{4x - x^3}{4} ]

Maximizing the Volume

To find the maximum volume, we take the derivative of (V) with respect to (x) and set it to zero:

[ V' 1 - frac{3x^2}{4} ]

Setting the derivative to zero:

[ 1 - frac{3x^2}{4} 0 implies frac{3x^2}{4} 1 implies x^2 frac{4}{3} implies x frac{2}{sqrt{3}} approx 1.155 ]

Calculating (h)

Substitute (x) back into the equation for (h):

[ h frac{4 - x^2}{4x} frac{4 - frac{4}{3}}{4 cdot frac{2}{sqrt{3}}} frac{frac{8}{3}}{frac{8}{sqrt{3}}} sqrt{3} ]

Calculating the Maximum Volume

Substitute (x) and (h) back into the volume formula:

[ V x^2h left(frac{4}{3}right) left(sqrt{3}right) frac{4sqrt{3}}{3} approx 2.309 , m^3 ]

Alternative Approach

Alternatively, we can also define the width (w x). The surface area constraint can be rewritten as:

[ S w^2 4wh 4 implies w frac{4 - w^2}{4w} implies V w^2 cdot frac{4 - w^2}{4w} frac{4w - w^3}{4} ]

To find the maximum volume, take the derivative and set it to zero:

[ V' 1 - frac{3w^2}{4} 0 implies w^2 frac{4}{3} implies w sqrt{frac{4}{3}} ]

Substitute (w) back into the volume equation:

[ V frac{4sqrt{frac{4}{3}} - sqrt{frac{4}{3}}^3}{4} frac{2}{3} sqrt{frac{4}{3}} frac{4sqrt{3}}{9} , m^3 ]

Conclusion

The maximum volume of the box, given the constraint of 4 m2 of material, is approximately 2.309 m3 for the first approach, or (frac{4sqrt{3}}{9} , m^3) for the alternative approach. This optimal design maximizes the use of material while giving the box the largest possible volume.