To solve this problem, we need to maximize the volume of a square-based prism with side length $x$ and height $h$, given a constraint on the surface area.

Step-by-Step Solution

1. Identify Given Information:

   • Volume of the prism: $V = x^2 h$.
   • Surface area constraint: $A = x^2 + 4xh = 300$ cm².

2. Express $h$ in Terms of $x$: From the surface area constraint: $x^2 + 4xh = 300$ Solving for $h$: $4xh = 300 - x^2$ $h = \frac{300 - x^2}{4x}$

3. Substitute $h$ into the Volume Formula: Substitute $h = \frac{300 - x^2}{4x}$ into the volume expression: $V = x^2 \cdot \frac{300 - x^2}{4x}$ Simplify: $V = \frac{300x - x^3}{4}$

4. Differentiate $V$ with Respect to $x$: To maximize $V$, take the derivative with respect to $x$ and set it to zero: $\frac{dV}{dx} = \frac{300 - 3x^2}{4}$ Set $\frac{dV}{dx} = 0$: $300 - 3x^2 = 0$ $3x^2 = 300$ $x^2 = 100$ $x = 10 \, \text{cm}$

5. Find $h$ Using $x = 10$: Substitute $x = 10$ into the expression for $h$: $h = \frac{300 - 10^2}{4 \times 10}$ $h = \frac{300 - 100}{40}$ $h = \frac{200}{40} = 5 \, \text{cm}$

Solution Summary

The dimensions of the square-based prism that maximize the volume are:

• $x = 10 \, \text{cm}$
• $h = 5 \, \text{cm}$

Would you like a detailed breakdown of each step or have further questions?

1. What are other methods to solve constrained optimization problems?
2. How does surface area constraint affect the volume maximization?
3. What if the surface area constraint was different—how would the solution change?
4. Can this method apply to non-square-based prisms?
5. How does calculus play a role in optimization problems?

Tip: In optimization problems with constraints, Lagrange multipliers are a useful method to consider.