Question

Crease-length problem A rectangular sheet of paper of width \(a\) and length \(b,\) where \(0

Short answer

Question: Given a rectangular piece of paper with width a and length b, find the point at which a fold will result in the minimum crease and determine the length of the minimum crease. Answer: Folding the paper at the point where DP = 0 will result in the minimum crease. The length of the minimum crease is given by \(L_{min} = \sqrt{2a^2 - b^2 + 2ab}\).

Step-by-step solution

Set up the problem

First, let's label some points. Let A, B, and C be the vertices of width a on the short side, and D, E, F be the vertices of length b on the long side (A and D are together). Let P be the point on segment DE at which the paper is folded, with G and H being the points on the paper that P is folded to touch. Now, we have a triangle, DPG, whose dimensions we have to find. Let \(DP = x, PG = y\).

Find the dimensions

Now, we have to use the Pythagorean theorem to find the dimensions of the triangle DPG. G is the point on segment BC that lies above P. So, we have two triangles - APG and ABC sharing a side AG. Therefore, we can write the equations: \(AP^2 = a^2 + x^2\) \(BP^2 = (b-a)^2 + y^2\) Also, in triangle APG and triangle ABC, we have: \(AG^2 + PG^2 = AC^2 = (x^2 + a^2) + (y^2 + (b-a)^2)\)

Optimize the crease function

Let L be the length of the crease. Then, \(L^2 = DP^2 + PG^2 = x^2 + y^2\) We want to minimize L. Using the equations of step 2, we get: \(L^2 = (a^2 + x^2) - (b-a)^2 = x^2 - (b-a)^2 + a^2\) Since \(L^2\) is a function of x (and not y), let's find the first and second derivatives of \(L^2 (x)\) with respect to x, and use them to optimize the function. First derivative: \(\frac{d}{dx}(L^2) = 2x\) Second derivative: \(\frac{d^2}{dx^2}(L^2) = 2\)

Analyze results and find the minimum crease

Since the second derivative is positive and constant, the function of \(L^2(x)\) is convex parabolic, and thus the minimum occurs at the critical points. At the critical points, the first derivative is equal to 0. \(2x = 0 \Rightarrow x = 0\) This means that the minimum crease occurs when DP = 0. Now, we have to find the length of the minimum crease.

Calculate the length of the minimum crease

Using the relationships we derived earlier: \(L^2_{min} = a^2 - (b-a)^2 + a^2 = 2a^2 - b^2 + 2ab\) So, the minimum length of the crease is: \(L_{min} = \sqrt{2a^2 - b^2 + 2ab}\)

Key concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry, and it plays a vital role in solving optimization problems in calculus. This theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, if we have a right triangle with sides of length 'a' and 'b', and hypotenuse of length 'c', then the relationship is given by the equation:
\[ c^2 = a^2 + b^2 \]
In the context of the crease-length problem, we apply this theorem to establish the relationship between the lengths of the triangle's sides. Understanding how to leverage this theorem allows students to convert a spatial problem into an algebraic one, which is then tackled using calculus methods for optimization.

First derivative test

The first derivative test is an essential tool in calculus, used to determine where a function has relative maxima or minima. It involves taking the derivative of the function with respect to its variable and finding the critical points where the derivative is zero or undefined. These points are then tested to see if they indicate a peak or a trough in the function's graph.

In our crease-length problem, the first derivative of the crease function \( L^2 (x) \) is calculated as \( 2x \). Setting this derivative equal to zero helps us find the critical points, which, in our case, suggest that there is a potential minimum at \( x = 0 \). Understanding the first derivative test is critical for students as it paves the way to find where the function may have optimal values—such as the shortest crease length in this scenario.

Second derivative test

The second derivative test complements the first derivative test by providing additional information about the nature of the critical points identified previously. Once the critical points are found, taking the second derivative of the function can tell us whether each point corresponds to a concave upward (indicating a minimum) or concave downward (indicating a maximum) section of the function's graph.

In our example, the second derivative of \( L^2(x) \) is a constant \( 2 \), signifying that the function is concave upward everywhere. Since a positive second derivative indicates that the function is convex or 'bowl-shaped', any critical point found by the first derivative test will be a minimum. This knowledge helps students conclude that the shortest possible crease is indeed located at the derived critical point.

Convex functions

Convex functions play a critical role when solving optimization problems in calculus. A function is considered convex on an interval if the line segment between any two points on the graph of the function lies above or on the graph over that interval. This property is especially useful because it guarantees that any local minimum is also a global minimum, simplifying the search for optimal solutions.

For the crease-length problem, we identified that \( L^2(x) \) is a convex function since its second derivative is positive. Convexity ensures that the crease of minimum length found at \( x = 0 \) is the shortest crease possible for the entire sheet. It's essential for students to grasp the concept of convex functions, as this understanding can significantly ease the process of optimization by narrowing down the possible solutions to one definitive answer.