Math Problem Statement

As shown, an open box has a length of (52-2x) inches, a width of (42-2x) inches, and a height of x inches. Describe how to create this box from a rectangular piece of cardboard, identify the degree of the polynomial that represents the volume, and describe a procedure for finding the value of x that yields the maximum possible volume.

Solution

The image depicts a problem involving an open-top box with a length of 522x52 - 2x inches, a width of 422x42 - 2x inches, and a height of xx inches. The problem requires solving several parts:

A. Describing how to make the box from a rectangular piece of cardboard:

To create the box, start with a rectangular piece of cardboard with original dimensions 5252 inches by 4242 inches. Cut out squares of side xx inches from each corner of the cardboard. After removing these squares, fold up the flaps created by cutting the squares to form the sides of the box. This creates the open-top box with dimensions (522x)×(422x)×x(52 - 2x) \times (42 - 2x) \times x.

B. Degree of the polynomial representing the volume:

The volume V(x)V(x) of the box is calculated by multiplying the length, width, and height: V(x)=(522x)(422x)(x)V(x) = (52 - 2x)(42 - 2x)(x) First, expand the product of the length and width: (522x)(422x)=2184188x+4x2(52 - 2x)(42 - 2x) = 2184 - 188x + 4x^2 Now, multiply by the height xx: V(x)=x(2184188x+4x2)=2184x188x2+4x3V(x) = x(2184 - 188x + 4x^2) = 2184x - 188x^2 + 4x^3 Thus, the volume is represented by the polynomial: V(x)=4x3188x2+2184xV(x) = 4x^3 - 188x^2 + 2184x The degree of this polynomial is 3 because the highest power of xx is x3x^3.

C. Procedure for finding the value of xx that maximizes the volume:

To find the value of xx that maximizes the volume, we need to find the critical points by taking the derivative of V(x)V(x), setting it equal to zero, and solving for xx.

The derivative of the volume function is: V(x)=12x2376x+2184V'(x) = 12x^2 - 376x + 2184 Set this equal to zero to find the critical points: 12x2376x+2184=012x^2 - 376x + 2184 = 0 Solve this quadratic equation using the quadratic formula: x=(376)±(376)24(12)(2184)2(12)x = \frac{-(-376) \pm \sqrt{(-376)^2 - 4(12)(2184)}}{2(12)} This will give the value of xx that maximizes the volume.

Let me know if you would like me to continue solving this quadratic equation!


Would you like further details or have any questions?

Here are 5 related questions:

  1. What are the possible values of xx based on the dimensions of the box?
  2. Why is the degree of the volume polynomial important?
  3. How would you interpret the critical points in the context of maximizing volume?
  4. Could there be a minimum volume for certain values of xx, or is it always positive?
  5. How does cutting squares from the cardboard influence the overall volume?

Tip: When solving optimization problems, always verify that the solution is within the problem's domain (i.e., check that the dimensions make sense physically).

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Math Problem Analysis

Mathematical Concepts

Polynomials
Optimization
Derivatives

Formulas

Volume of a box: V = length × width × height
Polynomial expansion: (a - bx)(c - dx)
Derivative of a polynomial: V'(x) = 12x^2 - 376x + 2184

Theorems

Quadratic Formula
Maximization using Derivatives

Suitable Grade Level

Grades 10-12