Optimizing Cylinder Dimensions for Minimal Metal Cost in Manufacturing

Introduction

In manufacturing cylindrical cans, efficiency and cost-effectiveness are paramount. This article delves into the use of calculus to determine the optimal dimensions of a cylindrical can to minimize the cost of the metal used. By understanding the mathematical principles behind this optimization, manufacturers can reduce material waste and production expenses.

Setting Up the Problem

Given that the volume of the cylinder must be 1000 cm3, we start by defining the volume and surface area formulas for a cylinder, which are essential to the optimization process.

Volume and Surface Area Formulas

The volume V of a cylinder is given by the formula:

[ V pi r^2 h ]

where r is the radius and h is the height of the cylinder. Given a volume of 1000 cm3, we have:

[ pi r^2 h 1000 ]

The surface area S of the cylinder, which directly relates to the cost of the metal, is given by:

[ S 2pi r^2 2pi rh ]

This consists of the areas of the two circular bases and the side surface area.

Expressing h in Terms of r

To proceed, we express h in terms of r using the volume equation:

[ h frac{1000}{pi r^2} ]

Substituting h into the Surface Area Formula

Substituting the expression for h into the surface area formula, we get:

[ S 2pi r^2 2pi r left( frac{1000}{pi r^2} right) ]

Simplifying this, we obtain:

[ S 2pi r^2 frac{2000}{r} ]

Minimizing the Surface Area

To find the minimum surface area, we differentiate S with respect to r and set the derivative to zero:

[ frac{dS}{dr} 4pi r - frac{2000}{r^2} ]

Setting the derivative equal to zero:

[ 4pi r - frac{2000}{r^2} 0 ]

Rearranging to solve for r:

[ 4pi r^3 2000 ] [ r^3 frac{500}{pi} ] [ r left( frac{500}{pi} right)^{1/3} ]

Approximating the value:

[ r approx left( frac{500}{3.14159} right)^{1/3} approx 5.419 text{ cm} ]

Substituting r back into the expression for h:

[ h frac{1000}{pi r^2} ]

Calculating h:

[ h approx frac{1000}{pi cdot 29.38} approx frac{1000}{92.57} approx 10.81 text{ cm} ]

Conclusion

The dimensions of the cylindrical can that minimize the cost of the metal are approximately:

Radius r: 5.42 cm Height h: 10.81 cm

These dimensions ensure a volume of 1000 cm3 while minimizing the surface area.

General Result

A more general result states that when the height of the can is twice the radius, the surface area is minimized. This implies:

[ h 2r ]

Substituting h 2r into the surface area formula:

[ A 2pi r h 2pi r^2 2pi r (2r) 2pi r^2 4pi r^2 2pi r^2 6pi r^2 ]

For a given volume V, we have:

[ 2pi r h V Rightarrow 2pi r (2r) V Rightarrow 4pi r^2 frac{V}{r} Rightarrow A frac{V}{r} 2pi r^2 ]

To minimize A, we differentiate and set the derivative to zero:

[ frac{dA}{dr} -frac{V}{r^2} 4pi r 0 Rightarrow 4pi r^3 V Rightarrow r left( frac{V}{4pi} right)^{1/3} ]

Therefore, the optimized height is:

[ h 2r 2 left( frac{V}{4pi} right)^{1/3} ]

Related Articles and Resources

For further reading, consult the following articles:

Calculus Optimization Problems: A collection of optimization problems using calculus to find the maximum or minimum values. Optimizing Volume and Surface Area in Engineering: Practical applications of optimization in engineering and manufacturing. Manufacturing Cost Analysis: Detailed analysis of how to minimize manufacturing costs through effective design and material optimization.