To minimize the amount of material used in constructing an open box with a square base and a given volume, we need to minimize the surface area subject to the constraint of a fixed volume. Here's how we can approach this:

Given:

• The volume of the box: $V = 103 \, \text{in}^3$
• The base is square with side length $x$, and the height of the box is $h$.

Step 1: Express the volume constraint

The volume $V$ of the box is given by: $V = x^2 h$ We know $V = 103$, so: $x^2 h = 103$ This gives us the relationship between $x$ and $h$: $h = \frac{103}{x^2}$

Step 2: Express the surface area

Since the box has no top, the surface area $A$ consists of the area of the base and the four sides. The surface area is: $A = x^2 + 4xh$ Substitute $h$ from the volume constraint into the surface area formula: $A = x^2 + 4x \left( \frac{103}{x^2} \right)$ Simplify this expression: $A = x^2 + \frac{412}{x}$

Step 3: Minimize the surface area

To minimize $A$, we take the derivative of $A$ with respect to $x$, set it equal to zero, and solve for $x$.

First, find the derivative: $\frac{dA}{dx} = 2x - \frac{412}{x^2}$ Set this equal to zero to find the critical points: $2x - \frac{412}{x^2} = 0$ Multiply both sides by $x^2$ to eliminate the fraction: $2x^3 = 412$ Solve for $x$: $x^3 = 206$ $x = \sqrt[3]{206} \approx 5.88 \, \text{in}$

Step 4: Find the height $h$

Now that we have $x$, we can find the height $h$ using the volume equation: $h = \frac{103}{x^2} = \frac{103}{(5.88)^2} \approx 2.98 \, \text{in}$

Final Dimensions:

• Height: $h \approx 2.98 \, \text{in}$
• Length: $x \approx 5.88 \, \text{in}$
• Width: $x \approx 5.88 \, \text{in}$

Would you like more details on any specific step?

Related questions:

1. How do we confirm that the critical point is a minimum?
2. How would the process change if the box had a lid?
3. How does the material cost change if the volume is increased?
4. What if the base of the box is rectangular, not square?
5. How would the dimensions change if the surface area were fixed instead of the volume?

Tip:

When minimizing or maximizing quantities in constrained optimization problems, always express one variable in terms of the other to reduce the number of variables in your equations.