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# The aim is to build a swimming pool with a volume of 32m³ and a square base. What dimensions minimize the amount of material used to cover the walls $sides and bottom$ of a pool.

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## Answer to a math question The aim is to build a swimming pool with a volume of 32m³ and a square base. What dimensions minimize the amount of material used to cover the walls $sides and bottom$ of a pool.

Hank
4.8
1. Let $x$ be the length of one side of the square base, and $h$ be the height of the pool.

2. Given that the volume $V$ is 32 cubic meters, we have:
x^2 \cdot h = 32
h = \frac{32}{x^2}

3. The surface area to be minimized includes the area of the bottom and the four sides:
A = x^2 + 4xh

4. Substitute $h$ from step 2 into the surface area formula:
A = x^2 + 4x \left$\frac{32}{x^2} \right$
A = x^2 + \frac{128}{x}

5. To find the minimum surface area, take the derivative of $A$ with respect to $x$ and set it to zero:
\frac{dA}{dx} = 2x - \frac{128}{x^2}
2x - \frac{128}{x^2} = 0

6. Solve for $x$:
2x^3 = 128
x^3 = 64
x = 4

7. Substitute $x = 4$ back into the equation for $h$:
h = \frac{32}{4^2}
h = \frac{32}{16}
h = 2

Thus, the dimensions that minimize the amount of material are:
\boxed{x = 4 \text{ meters}, \; h = 2 \text{ meters}}

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