To minimize the amount of paper used for the rectangular poster, we need to design a rectangle with a fixed area for printing, while considering the margins. The goal is to find the dimensions of the poster that minimize its total area.

Problem Breakdown:

• Area of the printed content = 75 square inches (in²)
• Margins:
  • 3 inches on the top and bottom
  • 1 inch on each side

Let the dimensions of the entire poster be:

• Width of the poster = $x$ (in inches)
• Height of the poster = $y$ (in inches)

Thus, the area of the printed content is the area of the interior rectangle (after accounting for margins):

• The printed width = $x - 2$ (since there are 1-inch margins on both sides).
• The printed height = $y - 6$ (since there are 3-inch margins on the top and bottom).

We are told that the area of the printed content is 75 in², so we can set up the following equation: $(x - 2)(y - 6) = 75$

Objective: Minimize the total area of the poster

The total area of the poster is $A = x \cdot y$.

We can solve this optimization problem by expressing one variable in terms of the other using the equation for the printed area and then minimizing the total area.

1. From $(x - 2)(y - 6) = 75$, solve for $y$: $y - 6 = \frac{75}{x - 2}$ $y = \frac{75}{x - 2} + 6$

2. Substitute this expression for $y$ into the total area formula: $A = x \cdot \left( \frac{75}{x - 2} + 6 \right)$ Simplify this expression: $A = \frac{75x}{x - 2} + 6x$

Minimize the area function:

To find the value of $x$ that minimizes the total area, we take the derivative of $A(x)$ with respect to $x$, set it equal to zero, and solve for $x$.

First, calculate the derivative: $\frac{dA}{dx} = \frac{d}{dx} \left( \frac{75x}{x - 2} + 6x \right)$

This derivative requires the quotient rule for the first term. Let's calculate it:

$\frac{d}{dx} \left( \frac{75x}{x - 2} \right) = \frac{(x - 2) \cdot 75 - 75x \cdot 1}{(x - 2)^2} = \frac{75x - 150 - 75x}{(x - 2)^2} = \frac{-150}{(x - 2)^2}$ $\frac{d}{dx}(6x) = 6$

Now combine these: $\frac{dA}{dx} = \frac{-150}{(x - 2)^2} + 6$

Set $\frac{dA}{dx} = 0$ to find the critical points: $\frac{-150}{(x - 2)^2} + 6 = 0$ $\frac{-150}{(x - 2)^2} = -6$ $\frac{150}{(x - 2)^2} = 6$ $(x - 2)^2 = \frac{150}{6} = 25$ $x - 2 = 5 \quad \text{or} \quad x - 2 = -5$ $x = 7 \quad \text{or} \quad x = -3$

Since $x$ must be positive, we take $x = 7$.

Find $y$:

Substitute $x = 7$ into the equation for $y$: $y = \frac{75}{x - 2} + 6 = \frac{75}{7 - 2} + 6 = \frac{75}{5} + 6 = 15 + 6 = 21$

Conclusion:

The dimensions of the poster that minimize the total area are:

• Width = 7 inches
• Height = 21 inches

These dimensions will minimize the amount of paper used.

---

Would you like further details on the optimization process or any other questions? Here are 5 related questions:

1. How do we confirm that this solution minimizes the area and isn't a maximum?
2. What if the margins changed (e.g., 2 inches on the sides)?
3. How would you handle this problem if the printed area was not fixed?
4. What is the role of the derivative in optimization problems like this?
5. How can we apply this optimization strategy to other real-world design problems?

Tip: When minimizing areas with fixed constraints, always use derivatives to find the critical points, and verify if they represent a minimum by checking the second derivative or using a test for concavity.