Questions: Los márgenes superior e inferior de un cartel son de 6 cm y los márgenes de los lados de 4 cm. Si el área de impresión sobre el cartel es de 384 cm², encuentre las dimensiones del cartel con la menor área.

△ Find the dimensions of the poster with the minimum area given fixed margins and printing area. ○ Define variables and constraints ☼ Let \( x \) be the width and \( y \) be the height of the printing area. The total width of the poster is \( x + 8 \) cm, and the total height is \( y + 12 \) cm. The printing area is \( x \times y = 384 \, \text{cm}^2 \). ○ Express area in terms of one variable ☼ The total area of the poster is \( A = (x + 8)(y + 12) \). Using \( y = \frac{384}{x} \), substitute to get \( A = (x + 8)\left(\frac{384}{x} + 12\right) = 480 + 12x + \frac{3072}{x} \). ○ Find the minimum area using calculus ☼ Differentiate \( A \) with respect to \( x \): \(\frac{dA}{dx} = 12 - \frac{3072}{x^2}\). Set \(\frac{dA}{dx} = 0\) to find critical points: \( 12 = \frac{3072}{x^2} \), so \( x^2 = 256 \) and \( x = 16 \). ○ Calculate corresponding dimensions ☼ With \( x = 16 \), find \( y = \frac{384}{16} = 24 \). The poster dimensions are width \( 16 + 8 = 24 \, \text{cm} \) and height \( 24 + 12 = 36 \, \text{cm} \). ○ Verify minimum condition ☼ The second derivative \(\frac{d^2A}{dx^2} = \frac{2 \times 3072}{x^3}\) is positive for \( x > 0 \), confirming a minimum. ✧ The dimensions of the poster with the minimum area are 24 cm × 36 cm, with a printing area of 16 cm × 24 cm. ☺ The dimensions of the poster with the minimum area are 24 cm × 36 cm, with a printing area of 16 cm × 24 cm.