The problem asks for the dimensions of a rectangular box with the largest possible volume, given that its total surface area is 64 cm².

Key Information:

1. Volume of the box: $V = x \cdot y \cdot z$, where $x, y,$ and $z$ are the dimensions.
2. Surface area constraint: $2(xy + xz + yz) = 64$.

Goal:

Maximize $V = x \cdot y \cdot z$ subject to the surface area constraint.

Let me solve this step by step:

Step 1: Express the constraint.

The surface area is: $2(xy + xz + yz) = 64 \implies xy + xz + yz = 32.$

Step 2: Use optimization techniques.

To maximize $V = x \cdot y \cdot z$, we will rewrite one variable in terms of the others using the constraint. Let’s express $z$ in terms of $x$ and $y$: $z = \frac{32 - xy}{x + y}.$

Substitute $z$ into the volume equation: $V = x \cdot y \cdot \frac{32 - xy}{x + y}.$

This is now a function $V(x, y)$ that can be maximized using calculus.

Step 3: Solve for critical points.

We need to take partial derivatives of $V(x, y)$ with respect to $x$ and $y$, set them to zero, and solve for $x$ and $y$. After finding $x$ and $y$, we calculate $z$ using the constraint.

Let me compute this systematically. sympy import symbols, solve, diff

Define variables

x, y, z = symbols('x y z')

Constraint: xy + xz + yz = 32

z_expr = (32 - x*y) / (x + y)

Volume: V = x * y * z

V = x * y * z_expr

Partial derivatives

dV_dx = diff(V, x) dV_dy = diff(V, y)

Solve for critical points where derivatives are zero

critical_points = solve([dV_dx, dV_dy], (x, y), dict=True) critical_pointsThe