1) A small rectangular pool has dimensions 12 m by 6 m by 2 m . A landscaper wishes to increase the length, width and height by the same amount so that the surface area is at most \( 440 \mathrm{~m}^{2} \). What are the maximum dimensions of the larger pool?
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To find the maximum dimensions of the larger pool, let's denote the amount by which the length, width, and height are increased as \( x \). The new dimensions of the pool would be \( (12 + x) \), \( (6 + x) \), and \( (2 + x) \). The surface area \( S \) of a rectangular pool is calculated using the formula: \[ S = 2lw + 2lh + 2wh \] Plugging the dimensions into the formula gives us: \[ S = 2(12 + x)(6 + x) + 2(12 + x)(2 + x) + 2(6 + x)(2 + x) \] Simplifying this equation: \[ S = 2((12 + x)(6 + x) + (12 + x)(2 + x) + (6 + x)(2 + x)) \] Expanding each term and combining like terms eventually leads to: \[ S = 2(72 + 18x + 12x + 6x + 24 + 8x + 12 + 4x) = 2(72 + 38x) = 144 + 76x \] Setting this less than or equal to 440 gives us: \[ 144 + 76x \leq 440 \] Solving for \( x \): \[ 76x \leq 440 - 144 \] \[ 76x \leq 296 \] \[ x \leq \frac{296}{76} \] \[ x \leq 3.89 \] (approximately) Thus, to find the maximum dimensions, we take \( x = 3.89 \) m, rounding to two decimal points gives us \( x \approx 3.89 \) m. Therefore, the maximum dimensions of the larger pool are: - Length: \( 12 + 3.89 \approx 15.89 \) m - Width: \( 6 + 3.89 \approx 9.89 \) m - Height: \( 2 + 3.89 \approx 5.89 \) m So, the maximum dimensions are approximately \( 15.89 \mathrm{~m} \times 9.89 \mathrm{~m} \times 5.89 \mathrm{~m} \).