To minimize the surface area of a rectangular box with an open top, given that the length $l$ is three times the width $w$ and the volume $V$ is 220, we can follow these steps:

1. Define the variables:

   • Let the width be $w$.
   • Then, the length $l = 3w$.
   • The height is $h$.

2. Volume equation: $V = l \times w \times h = 220$ Substituting for $l$: $3w \times w \times h = 220 \implies 3w^2h = 220 \implies h = \frac{220}{3w^2}$

3. Surface area equation: The surface area $S$ of the open box is given by: $S = lw + lh + wh = 3w \cdot w + 3w \cdot h + w \cdot h$ Substituting for $h$: $S = 3w^2 + 3w \cdot \frac{220}{3w^2} + w \cdot \frac{220}{3w^2}$ Simplifying this gives: $S = 3w^2 + \frac{220}{w} + \frac{220}{3w}$ Combining the last two terms: $S = 3w^2 + \frac{220 + \frac{220}{3}}{w} = 3w^2 + \frac{220 \cdot 4/3}{w} = 3w^2 + \frac{880}{3w}$

4. Minimize the surface area: To minimize $S$, we take the derivative and set it to zero: $\frac{dS}{dw} = 6w - \frac{880}{3w^2}$ Setting the derivative to zero: $6w - \frac{880}{3w^2} = 0$ $6w^3 = \frac{880}{3} \implies w^3 = \frac{880}{18} \implies w^3 = \frac{440}{9} \implies w = \left(\frac{440}{9}\right)^{1/3}$

   Calculating $w$: $w \approx 4.88 \quad (\text{approx})$

5. Calculate $l$ and $h$:

   • $l = 3w \approx 14.64$
   • $h = \frac{220}{3w^2} \approx 3.07$

6. Minimum surface area: Finally, substitute $w$ back into the surface area equation to find the minimum surface area.

Conclusion:

The dimensions of the box that minimize the surface area with the given constraints are approximately:

• Width $w \approx 4.88$
• Length $l \approx 14.64$
• Height $h \approx 3.07$

Calculating the exact minimum surface area would require substituting these values back into the surface area formula.

Would you like details on any of these steps or have any questions? Here are some related questions you might consider:

1. How does changing the volume affect the dimensions?
2. What would happen if the box had a closed top?
3. Can you derive the dimensions for a different volume?
4. How does the surface area formula change with a different shape?
5. What is the significance of the dimensions in practical applications?

Tip: When optimizing shapes, using derivatives can often reveal minimum or maximum values efficiently.