Maximizing the Volume of a Box Constructed from Zinc Sheet

In calculus, we often encounter problems where we need to maximize or minimize certain values within given constraints. A practical example involves creating a box from a piece of zinc by cutting equal squares from each corner and folding the edges to form the sides. This article will demonstrate how to find the dimensions that will yield the largest possible volume of the box.

Problem Definition

The problem is as follows: a rectangular zinc sheet measures 20 square inches. We need to cut out squares of equal size from each corner and fold the edges to form a box. What is the largest volume of the box that can be constructed?

Mathematical Formulation

Notation:

x - the side length of the squares cut from each corner. Length: 20 - 2x Width: 20 - 2x Height: x

The volume (V) of the box can be expressed as:

Volume Function

[ V (20 - 2x)(20 - 2x)x (20 - 2x)^2 x ]

Expanding the Volume Function

[ V (20 - 4x 4x^2)x 2 - 4x^2 4x^3 ]

Find the Critical Points

To maximize the volume, we take the derivative (frac{dV}{dx}) and set it to zero:

[ frac{dV}{dx} 12x^2 - 16(20 - 2x) 12x^2 - 320 32x ]

Setting the derivative equal to zero:

[ 12x^2 32x - 320 0 ]

Solving the Quadratic Equation

Using the quadratic formula (x frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a 12), (b 32), and (c -320):

[ x frac{-32 pm sqrt{32^2 - 4 cdot 12 cdot (-320)}}{2 cdot 12} frac{-32 pm sqrt{1024 15360}}{24} frac{-32 pm sqrt{16384}}{24} ]

[ x frac{-32 pm 128}{24} ] [ x frac{96}{24} 4 ; text{or} ; x frac{-160}{24} approx -6.67 ]

Since cutting out a square of side length 4 inches is feasible, we discard the negative solution.

Checking Feasibility and Calculating the Maximum Volume

Substituting (xapprox 3.33) back into the volume function:

[ V 20 - 2 cdot 3.33(20 - 2 cdot 3.33)3.33 20 - 6.6620 - 6.663.33 13.3413.343.33 ]

Therefore, the volume is approximately:

[ V approx 13.34^2 cdot 3.33 approx 178.56 cdot 3.33 approx 594.36 ; text{cubic inches} ]

Verifying the Maximum Volume

By applying the AM-GM inequality, we can verify the maximum volume:

[ sqrt[3]{4x(20-2x)^2} leq frac{4x 2(20-2x)}{3} frac{40}{3} ]

Therefore, the maximum volume is bounded by:

[ 4x(20-2x)^2 leq left( frac{40}{3} right)^3 frac{64000}{27} approx 594.36 ; text{cubic inches} ]

Conclusion

We have demonstrated that by cutting out squares of 3.33 inches from each corner, we can achieve the largest possible volume of the box, which is approximately 594.36 cubic inches. This problem showcases the application of calculus and the AM-GM inequality in optimization.