To determine the dimensions of a rectangular box with a volume of 32 ft³ that minimizes the surface area (and thus the material used), we can approach the problem using optimization techniques.

Step 1: Problem setup

The box has:

• Length $l$,
• Width $w$,
• Height $h$.

The box is open at the top, so it has only five surfaces: the bottom, two sides, and two ends.

Volume constraint:

The volume of the box is given by the formula:

$V = l \cdot w \cdot h = 32 \text{ ft}^3$

Surface area:

The surface area $A$ (which we want to minimize) includes:

• Bottom: $l \cdot w$,
• Two sides: $2 \cdot l \cdot h$,
• Two ends: $2 \cdot w \cdot h$.

Thus, the surface area is:

$A = l \cdot w + 2 \cdot l \cdot h + 2 \cdot w \cdot h$

Step 2: Express $h$ in terms of $l$ and $w$

From the volume equation:

$l \cdot w \cdot h = 32 \implies h = \frac{32}{l \cdot w}$

Step 3: Substitute $h$ into the surface area equation

Substitute $h = \frac{32}{l \cdot w}$ into the surface area formula:

$A = l \cdot w + 2 \cdot l \cdot \left( \frac{32}{l \cdot w} \right) + 2 \cdot w \cdot \left( \frac{32}{l \cdot w} \right)$

Simplifying:

$A = l \cdot w + \frac{64}{w} + \frac{64}{l}$

Step 4: Optimization

To minimize the surface area, we take the partial derivatives of $A$ with respect to $l$ and $w$, set them equal to zero, and solve for the dimensions of the box.

Partial derivative with respect to $l$:

$\frac{\partial A}{\partial l} = w - \frac{64}{l^2}$

Set this equal to zero:

$w = \frac{64}{l^2}$

Partial derivative with respect to $w$:

$\frac{\partial A}{\partial w} = l - \frac{64}{w^2}$

Set this equal to zero:

$l = \frac{64}{w^2}$

Step 5: Solve the system of equations

Now we solve the system of two equations:

1. $w = \frac{64}{l^2}$
2. $l = \frac{64}{w^2}$

Substitute $w = \frac{64}{l^2}$ into the second equation:

$l = \frac{64}{\left( \frac{64}{l^2} \right)^2}$

This simplifies to:

$l = \frac{64 \cdot l^4}{64^2} \implies l^5 = 64^2 \implies l = \sqrt[5]{64^2} = 4$

Now that we have $l = 4$, substitute this back into $w = \frac{64}{l^2}$:

$w = \frac{64}{4^2} = \frac{64}{16} = 4$

Finally, substitute $l = 4$ and $w = 4$ into the volume equation to find $h$:

$h = \frac{32}{4 \cdot 4} = \frac{32}{16} = 2$

Step 6: Conclusion

The dimensions of the box that minimize the surface area are:

• Length $l = 4$ ft,
• Width $w = 4$ ft,
• Height $h = 2$ ft.

Summary

The optimal dimensions of the rectangular box are $4 \times 4 \times 2$ feet, giving a volume of 32 ft³ and minimizing the material used for construction.

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Would you like further explanation or details about any step? Here are 5 related questions for you:

1. How can we verify that these dimensions truly minimize the surface area?
2. What happens if the box is closed on top as well—how would the surface area change?
3. What is the general approach to solving optimization problems involving volume constraints?
4. How do we find the second derivative to confirm that this solution is a minimum?
5. How could this approach be applied to other optimization problems, such as minimizing cost for a cylindrical container?

Tip: When dealing with optimization problems like this, always start by writing down the constraints (like the volume) and the objective (like surface area) before applying derivatives.