Problem 31 - Calculus with Concepts in Calculus

Question

A rectangular printed page is to have margins 2 inches wide at the top and the bottom and margins 1 inch wide on each of the two sides. If the page is to have 35 square inches of printing, determine the minimum possible area of the page itself.

Step-by-Step Solution

Answer

The minimum possible area of the page is approximately 93.54 square inches.

Step 1: Understanding the Problem

We are given a rectangular printed page with specific dimensions for margins and a printed area of 35 square inches. We need to find the minimum total area of the page, considering the margins.

Step 2: Setting Up Variables

Let the width of the printed area be \( x \) inches and the height to be \( y \) inches. The printed area is then \( x \, y = 35 \) square inches. The total width of the page, including the margins, is \( x + 2 \) inches, and the total height is \( y + 4 \) inches.

Step 3: Expressing the Total Area

The total area of the page, including the margins, is given by \( (x + 2)(y + 4) \). We want to minimize this expression.

Step 4: Substituting the Printed Area Constraint

To use the relationship between \( x \) and \( y \), substitute \( y = \frac{35}{x} \) (since \( x \cdot y = 35 \)) into the total area expression: \[ A(x) = (x + 2)\left(\frac{35}{x} + 4\right) \].

Step 5: Simplifying the Total Area

Simplify the expression for the total area as: \[ A(x) = (x + 2) \left(\frac{35}{x} + 4\right) = 35 + 4x + \frac{70}{x} + 8 \],which simplifies further to \[ A(x) = 43 + 4x + \frac{70}{x} \].

Step 6: Finding the Derivative

Find the derivative of the area function with respect to \( x \). The derivative \( A'(x) \) is given by:\[ A'(x) = 4 - \frac{70}{x^2} \].

Step 7: Solving for Critical Points

Set the derivative equal to zero to find critical points:\[ 4 - \frac{70}{x^2} = 0 \]. Solving gives \( x^2 = \frac{70}{4} = 17.5 \), so \( x = \sqrt{17.5} \approx 4.18 \).

Step 8: Verifying Minimum Area

Verify that this value for \( x \) gives a minimum by checking the second derivative, \( A''(x) = \frac{140}{x^3} \), which is positive for all valid \( x \) values, confirming we have a minimum.

Step 9: Calculate the Minimum Area

Substitute \( x = \sqrt{17.5} \approx 4.18 \) back into the area function to find the minimum area. Calculate \( y = \frac{35}{x} \approx 8.37 \). The minimum area of the page is approximately \[ A(4.18) = (4.18 + 2)(8.37 + 4) \approx 93.54 \text{ square inches} \].

Key Concepts

Derivatives

Derivatives are a fundamental tool in calculus, especially when it comes to optimization problems. They help us determine how a function changes as its input changes. In the context of minimizing or maximizing a quantity, the derivative is used to identify critical points where these extremes may occur. To find the derivative of a function, you need to apply rules such as the product rule, quotient rule, or chain rule depending on the form of the function. In this particular problem, we differentiated the function representing the total area of a page relative to its width. This derivative allows us to observe how changes in width affect the total area, an essential step towards finding the minimum area.

Critical Points

Critical points are specific values of the input variable where the first derivative of a function is zero or undefined. They are important because they indicate potential locations of maximum or minimum values of the function. In our problem, we found the critical point by setting the derivative of the area function to zero. This led us to find that the critical width of the printed area should be approximately 4.18 inches. This critical point needs to be assessed further to ascertain whether it gives a minimum or a maximum area.

Second Derivative Test

The second derivative test is a powerful method to verify whether a critical point is a local maximum, minimum, or a point of inflection. This involves computing the second derivative of the function and evaluating it at the critical points. If the second derivative is positive, the function has a local minimum at that point, whereas if it is negative, the function has a local maximum. For the rectangular page problem, we computed the second derivative which turned out to be positive for the critical width calculation. This confirmed that the critical width indeed provided a local minimum for the area of the page.

Geometry in Calculus

Geometry plays a crucial role in solutions involving calculus, especially in optimization problems where spatial constraints are considered. The rectangular page problem incorporated geometric considerations such as margins on the sides of the page. Using geometry, the dimensions of the printed area were related to the total width and height of the page. Additionally, geometry helped in setting up the relationship between the printed area and total area, allowing us to express the problem in terms of calculable variables, such as width and height. This geometric setup is what ultimately allowed the use of calculus tools like derivatives and critical points to find the optimal solution.