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Critical points are essential in determining the maximum or minimum values of a function. These are the values of the variable where the derivative of the function equals zero. In simpler terms, critical points suggest locations where the function stops increasing or decreasing and may change direction.
For our box volume problem, after finding the polynomial, we calculated the derivative as \( V'(x) = 12x^2 - 104x + 160 \). By setting this derivative equal to zero, \( V'(x) = 0 \), we identified critical points, which are potential candidates for extreme values (maximum or minimum) of the volume. Always remember to check if these points are within the given domain. Here, it led us to evaluate \( x = 2 \) as a critical point within the specified range \( 0 < x < 5 \).
Evaluating critical points, along with checking endpoints, helps in forming a complete understanding of the behavior of the function across the domain.

The derivative of a function reveals the rate at which the function's value is changing at any given point. In our context, the derivative calculation is crucial for identifying points where the volume reaches an extreme value—either a maximum or a minimum.
Once the function \( V(x) = 4x^3 - 52x^2 + 160x \) was established, the next step involved finding its derivative. Differentiation simplifies finding changes in volume corresponding to changes in dimensions. By applying derivative rules, we determined \( V'(x) = 12x^2 - 104x + 160 \).
Understanding derivatives helps in analyzing functions' slopes at various points. A zero slope (derivative) indicates potential critical points where the function may showcase its maximum height, especially when considered along with endpoints.

Solving quadratic equations is made straightforward using the quadratic formula. This universal tool allows us to find exact solutions for any quadratic equation in the form \( ax^2 + bx + c = 0 \).
In our exercise, we needed to solve \( 12x^2 - 104x + 160 = 0 \) to identify critical points. The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) was applied with \( a = 12 \), \( b = -104 \), and \( c = 160 \). This calculation led to critical point solutions: \( x = 6.67 \) and \( x = 2 \). Importantly, we confirmed that only \( x = 2 \) was within the permissible domain \( 0 < x < 5 \).
Mastering the quadratic formula empowers you to solve for unknowns in quadratic scenarios, an essential skill in calculus and beyond.

Polynomial expansion refers to rewriting a product of expressions as a sum of terms involving powers of variables. For function \( V(x) \), the task was to expand \((10 - 2x)(16 - 2x)\) to simplify calculations for determining volume.
The expanded form, \( 160 - 32x - 20x + 4x^2 = 160 - 52x + 4x^2 \), was inserted back into the function. Further expansion yielded a cubic polynomial \( V(x) = 4x^3 - 52x^2 + 160x \).
This form containing all powers of \( x \) alongside their coefficients, simplifies the derivative computation and identification of important points for volume evaluation. Understanding these expansions is key because they provide clearer pathways for performing further mathematical operations like differentiation.