To minimize the amount of material used for constructing an open box with a square base and a given volume, we need to minimize the surface area. Let’s define the variables:

• Let $x$ be the length of the side of the square base (in inches).
• Let $h$ be the height of the box (in inches).

The volume of the box is given by: $V = x^2h$ We are told that the volume is 94 in³, so: $x^2h = 94$ From this equation, we can solve for $h$: $h = \frac{94}{x^2}$

The surface area $A$ of the open box consists of the area of the base and the area of the four sides (no top). The surface area is: $A = x^2 + 4xh$ Substitute $h = \frac{94}{x^2}$ into the surface area equation: $A = x^2 + 4x \left(\frac{94}{x^2}\right) = x^2 + \frac{376}{x}$

Now, we minimize the surface area by taking the derivative of $A$ with respect to $x$, and setting it equal to zero: $\frac{dA}{dx} = 2x - \frac{376}{x^2}$ Set this derivative equal to zero to find the critical points: $2x - \frac{376}{x^2} = 0$ Multiply through by $x^2$ to eliminate the denominator: $2x^3 = 376$ Solve for $x$: $x^3 = 188 \quad \Rightarrow \quad x = \sqrt[3]{188} \approx 5.76$

Now, find the height $h$ using $h = \frac{94}{x^2}$: $h = \frac{94}{(5.76)^2} \approx \frac{94}{33.18} \approx 2.83$

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

• Side length of the square base: $x \approx 5.76$ inches
• Height: $h \approx 2.83$ inches

Would you like more details on any step or have any questions? Here are five related questions to consider:

1. How does the concept of optimization apply to real-world problems?
2. What is the significance of using derivatives to find minimum or maximum values?
3. Can you explain the relationship between the volume and surface area of 3D objects?
4. How do we confirm that the critical point found is indeed a minimum and not a maximum?
5. What would change in this problem if the box had a lid?

Tip: Always verify if a critical point leads to a minimum or maximum by checking the second derivative or using a sign analysis.