To minimize the area of the poster while keeping the printed material area fixed, we need to use optimization techniques.

Problem Setup:

1. Let the width of the printed area be $x$ and the height of the printed area be $y$.
2. The area of the printed material is fixed:
   $x \cdot y = 382$
3. The total width of the poster is $x + 12$ (6 cm margins on both sides).
4. The total height of the poster is $y + 8$ (4 cm margins on top and bottom).

The total area of the poster is:
$A = (x + 12)(y + 8)$

Our goal is to minimize $A$, subject to the constraint $x \cdot y = 382$.

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Step 1: Express $y$ in terms of $x$:

From the constraint $x \cdot y = 382$, we get:
$y = \frac{382}{x}$

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Step 2: Substitute $y$ into $A$:

Substitute $y = \frac{382}{x}$ into the area formula:
$A = (x + 12) \left(\frac{382}{x} + 8\right)$ Simplify:
$A = (x + 12) \left(\frac{382}{x}\right) + (x + 12)(8)$ $A = 382 + \frac{4584}{x} + 8x + 96$ $A = 478 + \frac{4584}{x} + 8x$

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Step 3: Minimize $A$:

To find the minimum, take the derivative of $A$ with respect to $x$ and set it equal to 0:
$\frac{dA}{dx} = -\frac{4584}{x^2} + 8$ Set $\frac{dA}{dx} = 0$:
$-\frac{4584}{x^2} + 8 = 0$ $\frac{4584}{x^2} = 8$ $x^2 = \frac{4584}{8} = 573$ $x = \sqrt{573} \approx 23.95$

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Step 4: Find $y$:

Using $y = \frac{382}{x}$:
$y = \frac{382}{23.95} \approx 15.95$

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Step 5: Calculate dimensions of the poster:

• Total width = $x + 12 = 23.95 + 12 = 35.95$
• Total height = $y + 8 = 15.95 + 8 = 23.95$

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Final Answer:

The dimensions of the poster with the smallest area are approximately: $\text{Width} = 35.95 \, \text{cm}, \, \text{Height} = 23.95 \, \text{cm}.$

Would you like detailed calculations or help visualizing this?

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Related Questions:

1. How do we confirm the dimensions truly minimize the area?
2. What happens if the margins change to different values?
3. How is the concept of optimization used in other real-life problems?
4. Could we solve this using numerical methods instead of calculus?
5. How does the fixed printed area constraint affect the solution?

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Tip:

Always verify optimization problems by checking the second derivative or using boundary conditions. This ensures the solution is a true minimum or maximum.