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?

To find the maximum dimensions of the larger pool while ensuring the surface area does not exceed \(440 \, \mathrm{m}^{2}\), we first calculate the surface area of the existing pool. The formula for the surface area \(A\) of a rectangular pool is given by: \[ A = 2lw + 2lh + 2wh \] For the original pool dimensions (length \(l = 12 \, \mathrm{m}\), width \(w = 6 \, \mathrm{m}\), height \(h = 2 \, \mathrm{m}\)), the surface area is: \[ A = 2(12)(6) + 2(12)(2) + 2(6)(2) = 144 + 48 + 24 = 216 \, \mathrm{m}^{2} \] Let \(x\) be the increase in length, width, and height. The new dimensions will be: - Length: \(12 + x\) - Width: \(6 + x\) - Height: \(2 + x\) The new surface area can be described as: \[ A' = 2((12 + x)(6 + x) + (12 + x)(2 + x) + (6 + x)(2 + x)) \] We need this new surface area \(A'\) to be at most \(440 \, \mathrm{m}^{2}\): \[ 2((12 + x)(6 + x) + (12 + x)(2 + x) + (6 + x)(2 + x)) \leq 440 \] Dividing by 2: \[ (12 + x)(6 + x) + (12 + x)(2 + x) + (6 + x)(2 + x) \leq 220 \] Now, we will expand and simplify each term: 1. \( (12 + x)(6 + x) = 72 + 18x + x^2 \) 2. \( (12 + x)(2 + x) = 24 + 14x + x^2 \) 3. \( (6 + x)(2 + x) = 12 + 8x + x^2 \) Putting it all together: \[ (72 + 18x + x^2) + (24 + 14x + x^2) + (12 + 8x + x^2) \leq 220 \] Combining like terms: \[ 108 + 40x + 3x^2 \leq 220 \] Subtracting 220 from both sides: \[ 3x^2 + 40x - 112 \leq 0 \] Now, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = 40\), and \(c = -112\): Calculating the discriminant: \[ b^2 - 4ac = 40^2 - 4 \cdot 3 \cdot (-112) = 1600 + 1344 = 2944 \] Finding the roots: \[ x = \frac{-40 \pm \sqrt{2944}}{6} \approx \frac{-40 \pm 54.24}{6} \] Calculating the two roots: 1. \(x \approx \frac{14.24}{6} \approx 2.37\) 2. \(x \approx \frac{-94.24}{6} \approx -15.71\) (not applicable) Thus, \(x \leq 2.37\). Now substituting back: - Maximum length: \(12 + 2.37 \approx 14.37 \, \mathrm{m}\) - Maximum width: \(6 + 2.37 \approx 8.37 \, \mathrm{m}\) - Maximum height: \(2 + 2.37 \approx 4.37 \, \mathrm{m}\) The maximum dimensions of the larger pool are approximately: **Length: 14.37 m, Width: 8.37 m, Height: 4.37 m.**