Let's analyze the problem step-by-step.

Problem Setup:

1. Box Characteristics:

   • It has a square base with side length $x$.
   • The height of the box is $y$.
   • The box has an open top.
   • Volume $V = 78732 \, \text{cm}^3$.

2. Goal:

   • Minimize the surface area to find the dimensions of the box that require the least material.

Solution Breakdown:

1. Volume Formula:

   • Since $V = x^2 y$, we can solve for $y$ in terms of $x$ and $V$: $y = \frac{V}{x^2} = \frac{78732}{x^2}$

2. Surface Area Formula:

   • The surface area $A(x)$ of the box (with an open top) is given by: $A(x) = x^2 + 4xy$
   • Substitute $y$ from the volume equation: $A(x) = x^2 + 4x \left(\frac{78732}{x^2}\right) = x^2 + \frac{314928}{x}$
   • This matches the given answer.

3. Derivative of $A(x)$:

   • To minimize the surface area, we take the derivative $A'(x)$ with respect to $x$: $A'(x) = 2x - \frac{314928}{x^2}$
   • This derivative is also confirmed to be correct in the image.

4. Critical Point:

   • Set $A'(x) = 0$ to find the critical value of $x$: $2x - \frac{314928}{x^2} = 0$
   • Solve this equation for $x$ to find the value that minimizes $A(x)$.

Would you like a detailed solution for finding this critical value or have any questions?

Related Questions

1. What is the process of finding the minimum or maximum value of a function?
2. How can you verify if a critical point results in a minimum or maximum?
3. Why do we use derivatives for optimization in calculus?
4. How does the volume constraint affect the box’s dimensions?
5. What would happen if the box had a closed top instead?

Tip:

Always check that you substitute values correctly when using constraints to minimize or maximize functions.