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.

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Hank

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78 Answers

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}}

2. Given that the volume \( V \) is 32 cubic meters, we have:

3. The surface area to be minimized includes the area of the bottom and the four sides:

4. Substitute \( h \) from step 2 into the surface area formula:

5. To find the minimum surface area, take the derivative of \( A \) with respect to \( x \) and set it to zero:

6. Solve for \( x \):

7. Substitute \( x = 4 \) back into the equation for \( h \):

Thus, the dimensions that minimize the amount of material are:

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