Optimizing Wire Cut for Minimum Enclosed Area

Optimizing Wire Cut for Minimum Enclosed Area

In this article, we will explore a common optimization problem in calculus: determining how a wire of a given length should be cut to achieve the minimum total area enclosed by the shapes formed from each part of the wire. This type of problem can be solved using the techniques of calculus to find the critical points and subsequently the minimum value of the function.

We are given a piece of wire of length 1 meter. This wire is to be cut into two parts: one part is bent into the shape of a circle, and the other part is bent into the shape of a square. Our goal is to find the optimal cut that minimizes the total enclosed area.

Step 1: Express the Areas

Let x be the length of the wire used for the circle. Hence, the length of the wire used for the square is 1 - x.

The circumference of the circle is x. The radius of the circle, r, is given by r frac{x}{2pi}. The area of the circle, A_c, is A_c pi r^2 pi left(frac{x}{2pi}right)^2 frac{x^2}{4pi}. The perimeter of the square is 1 - x. The side length of the square, s, is s frac{1 - x}{4}. The area of the square, A_s, is A_s s^2 left(frac{1 - x}{4}right)^2 frac{(1 - x)^2}{16}.

Step 2: Total Area

The total area, A, is the sum of the areas of the circle and the square:

A A_c A_s frac{x^2}{4pi} frac{(1 - x)^2}{16}.

Step 3: Differentiate and Find Critical Points

To minimize the total area, we differentiate A with respect to x and set the derivative to zero:

frac{dA}{dx} frac{1}{2pi}x - frac{1 - x}{8}.

Multiplying through by 8pi to eliminate fractions:

4x - pi(1 - x) 0.

Simplifying:

4x - pi pi x 0,

x(4 pi) pi.

x frac{pi}{4 pi}.

Step 4: Determine the Length for the Square

The length of the wire used for the square is:

1 - x 1 - frac{pi}{4 pi} frac{4 pi - pi}{4 pi} frac{4}{4 pi}.

Step 5: Summary

The optimal cut can be summarized as:

x frac{pi}{4 pi} meters for the circle. 1 - x frac{4}{4 pi} meters for the square.

Application to the Second Problem

For the second problem, we have an expression for the area involving a circle and a square:

A pi r^2 s^2 - k(2pi r 4s) - 100.

Given:

frac{dA}{dr} 2pi r - 2pk 0 implies r frac{k}{pi}. frac{dA}{ds} 2s - 4k 0 implies s 2k. Substituting 4pi r 8s 100 to find k:

4pi left(frac{k}{pi}right) 8(2k) 100 implies 4k 16k 100 implies 20k 100 implies k 5

Therefore, r frac{5}{pi}, s 10.

The circle perimeter and square area are:

Circle length 100pi / (4pi) 25 cm. Square length 400pi / (4pi) 100 cm.

Minimum area calculation:

A pi left(frac{5}{pi}right)^2 10^2 - 5(2pi cdot frac{5}{pi} 4 cdot 10) - 100 frac{25}{pi} 100 - 100 - 100 frac{25}{pi} approx 7.96 cm^2.

Conclusion

This detailed step-by-step solution demonstrates the application of calculus to real-world problems, optimizing the use of materials to achieve minimum enclosed area. This method is useful in various fields, including engineering, architecture, and manufacturing.