## Step : **Form the Constraint and** Let the of the box \( l \) (length), \( w \) (width), and \( h \) (height). - **Girth** = \( 2w + 2h \) - **Constraint**: \( l + 2w + 2h = 130 \) - **Volume**: \( V = lwh \) **Explanation:** We write the constraint as an equation relating length and the sum of widths and heights, and set up the function to be maximized (the box's volume). --- ## Step 2: **Use Lagrange Multipliers and Symmetry** To maximize \( V = lwh \) subject to \( l + 2w + 2h = 130 \), for maximum volume in a rectangular box, it is optimal when \( w = h \) (by symmetry). So, let \( w = h = x \): - Constraint: \( l + 4x = 130 \implies l = 130 - 4x \) - Volume: \( V = l x^2 = (130 - 4x)x^2 \) To maximize, set the derivative to zero: \[ V = (130 - 4x)x^2 = 130x^2 - 4x^3 \] \[ \frac{dV}{dx} = 260x - 12x^2 = \] \[ 260x = 12x^2 \implies x = \frac{260}{12} = \frac{65}{3} \] \[ l = 130 - 4x = 130 - 4\left(\frac{65}{3}\right) = 130 - \frac{260}{3} = \frac{130 \cdot 3 - 260}{3} = \frac{390 - 260}{3} = \frac{130}{3} \] **Explanation:** We substitute \( w = h \) for simplicity, rewrite the constraint and volume in terms of a single variable, and find the critical point by setting the derivative equal to zero. --- ## Final Answer ### The largest volume that can be sent in a rectangular box is: \[ V = l x^2 = \left(\frac{130}{3}\right)\left(\frac{65}{3}\right)^2 = \frac{130}{3} \cdot \frac{4225}{9} = \frac{130 \times 4225}{27} = \frac{549250}{27} \approx 20342.6 \text{ cubic inches} \]

# Step-by-Step Solution to Maximize Volume of a Rectangular Box ## Step 1: **Formulate the Constraint and Variables** Let: - \( l \) = length of the box - \( w \) = width of the box - \( h \) = height of the box ### Definitions: - **Girth**: The girth is defined as \( 2w + 2h \). - **Constraint**: The post office restricts the package such that: \[ l + 2w + 2h = 130 \] - **Volume**: The volume \( V \) of the box is given by: \[ V = lwh \] **Explanation:** We establish the relationship between the dimensions of the box using the constraint equation which includes the length, width, and height. The volume function is set up to be maximized. --- ## Step 2: **Apply Lagrange Multipliers and Use Symmetry** To maximize the volume \( V = lwh \) under the constraint \( l + 2w + 2h = 130 \), we can simplify the problem using symmetry. Assume \( w = h = x \). ### New Variables: - Substitute \( h \) with \( x \): - Therefore, the constraint becomes: \[ l + 4x = 130 \implies l = 130 - 4x \] ### Volume Expression: - Now, express the volume \( V \) in terms of \( x \): \[ V = l \cdot w \cdot h = (130 - 4x)x^2 \] \[ V = 130x^2 - 4x^3 \] ### Maximizing Volume: To find the maximum volume, differentiate \( V \) with respect to \( x \): \[ \frac{dV}{dx} = 260x - 12x^2 \] Set the derivative to zero to find critical points: \[ 260x - 12x^2 = 0 \] Factoring out \( x \): \[ x(260 - 12x) = 0 \] This yields: \[ x = 0 \quad \text{or} \quad 260 - 12x = 0 \implies x = \frac{260}{12} = \frac{65}{3} \] ### Finding Length \( l \): Substitute \( x \) back to find \( l \): \[ l = 130 - 4x = 130 - 4\left(\frac{65}{3}\right) = 130 - \frac{260}{3} = \frac{130 \cdot 3 - 260}{3} = \frac{390 - 260}{3} = \frac{130}{3} \] **Explanation:** By assuming \( w = h \), we simplify the problem, express everything in terms of one variable, and then find the critical point where the volume is maximized. --- ## Final Answer ### The maximum volume that can be sent in a rectangular box is: \[ V = l \cdot x^2 = \left(\frac{130}{3}\right)\left(\frac{65}{3}\right)^2 = \frac{130}{3} \cdot \frac{4225}{9} = \frac{130 \cdot 4225}{27} = \frac{549250}{27} \approx 20342.6 \text{ cubic inches} \]