1. A rectangular page is to contain 24 square inches of print. The margins at the top and the bottom of the page are 1.5 inches, and the margins on the left and the right are 1 inch (see figures). What should be the measure of the page for the least amount of paper?

To minimize the amount of paper needed for the page, we need to minimize the total area of the rectangular page, given the constraint that the printed area is 24 square inches. Here's how we can approach this:

### Step-by-Step Solution:

1. **Define Variables:**
   Let the dimensions of the printed area be:
   - \( x \) = width of the printed area (in inches),
   - \( y \) = height of the printed area (in inches).

   We are told that the area of the printed region is 24 square inches:
   \[
   x \cdot y = 24 \quad \text{(Equation 1)}
   \]

2. **Dimensions of the Entire Page:**
   The total dimensions of the page will include the margins. The margins are:
   - 1 inch on both the left and right sides,
   - 1.5 inches on both the top and bottom.

   Therefore, the total dimensions of the page are:
   - Width of the page = \( x + 2 \) (since there are 1-inch margins on both sides),
   - Height of the page = \( y + 3 \) (since there are 1.5-inch margins on both the top and bottom).

   Thus, the total area \( A \) of the page is:
   \[
   A = (x + 2)(y + 3) \quad \text{(Equation 2)}
   \]

3. **Express \( y \) in Terms of \( x \):**
   From Equation 1, we know that \( x \cdot y = 24 \), so we can solve for \( y \):
   \[
   y = \frac{24}{x}
   \]

4. **Substitute \( y \) in the Area Formula:**
   Now, substitute \( y = \frac{24}{x} \) into Equation 2 to express the total area \( A \) in terms of \( x \) alone:
   \[
   A = (x + 2)\left(\frac{24}{x} + 3\right)
   \]

5. **Simplify the Expression:**
   Expand and simplify the expression for the area:
   \[
   A = (x + 2)\left(\frac{24}{x} + 3\right)
   \]
   \[
   A = (x + 2) \left(\frac{24 + 3x}{x}\right)
   \]
   \[
   A = \frac{(x+2)(24+3x)}{x}
   \]
   Expand:
   \[
   A = \frac{24x + 48 + 3x^2 + 6x}{x}
   \]
   \[
   A = 24 + \frac{48}{x} + 3x + 6
   \]
   Simplify further:
   \[
   A = 3x + \frac{48}{x} + 30
   \]

6. **Minimize the Area:**
   To minimize the area, take the derivative of \( A \) with respect to \( x \) and set it equal to zero:
   \[
   \frac{dA}{dx} = 3 - \frac{48}{x^2}
   \]
   Set \( \frac{dA}{dx} = 0 \) to find the critical points:
   \[
   3 - \frac{48}{x^2} = 0
   \]
   \[
   \frac{48}{x^2} = 3
   \]
   \[
   x^2 = \frac{48}{3} = 16
   \]
   \[
   x = 4
   \]

7. **Find the Corresponding Value of \( y \):**
   From Equation 1, we know that \( x \cdot y = 24 \). Substituting \( x = 4 \):
   \[
   4 \cdot y = 24
   \]
   \[
   y = 6
   \]

8. **Final Dimensions of the Page:**
   The total width of the page is:
   \[
   x + 2 = 4 + 2 = 6 \text{ inches}
   \]
   The total height of the page is:
   \[
   y + 3 = 6 + 3 = 9 \text{ inches}
   \]

Thus, the dimensions of the page that minimize the amount of paper used are **6 inches by 9 inches**.

---

Would you like further details or have any questions on this solution?

Here are 5 related questions for further exploration:

1. How would the solution change if the area of the printed region was different (e.g., 30 square inches)?
2. What is the effect on the page dimensions if the margins are increased or decreased?
3. How can you apply the method used here to minimize other quantities, like perimeter?
4. How can this concept be extended to pages with circular printed areas?
5. How would we solve the problem if the margins were not constant (e.g., variable top and bottom margins)?

**Tip:** When minimizing or maximizing areas, look for constraints and use derivatives to find critical points where the function has a minimum or maximum.

A rectangular page is to contain 24 square inches of print. The margins at the top and the bottom of the page are 1.5 inches, and the margins on the left and the right are 1 inch. What should be the measure of the page for the least amount of paper?

Discover how to optimize the size of a rectangular page that contains 24 square inches of print while minimizing the total paper area. The problem involves algebra, geometry, and calculus techniques, including taking derivatives to find critical points. Suitable for high school students in Grades 11-12 studying optimization problems.

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