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Chapter 12: Problem 41

Chocolate Box Company is going to make open-topped boxes out of \(6 \times 16\) -inch rectangles of cardboard by cutting squares out of the corners and folding up the sides. What is the largest volume box it can make this way?

Short Answer

Expert verified
The largest volume box that can be made from a 6x16 inch rectangle of cardboard by cutting squares out of the corners and folding up the sides has dimensions of \(12\times2\times2\) inches and a volume of \(48\) cubic inches.

Step by step solution

01

Assigning variables to the dimensions of the box

Let's assign variable x to the side length of the squares cut out of each corner of the cardboard. The remaining dimensions of the box would then be: - Length: 16-2x (since two corners are cut from both sides) - Width: 6-2x (since two corners are cut from both sides) - Height: x (the height will be equal to the length of the square cut from the corners)
02

Setting up the volume function

The volume of the box V, can be expressed as the product of its length, width, and height: V = (16-2x)(6-2x)(x)
03

Expanding the volume function

Expand the volume function V: V(x) = 4x^3 - 44x^2 + 96x
04

Find the critical points of the volume function

To find the critical points, take the first derivative of the volume function V'(x) and set it to zero: V'(x) = 12x^2 - 88x + 96 Solve for x: 12x^2 - 88x + 96 = 0 Divide by 4: 3x^2 - 22x + 24 = 0 Factor: (x - 2)(3x - 12) = 0 Solutions: x = 2 or x = 4
05

Determine which critical point yields the maximum volume

Take the second derivative of the volume function V''(x): V''(x) = 24x - 88 Evaluate V''(x) at x = 2: V''(2) = 24(2) - 88 = -40 (Since it's negative value, x = 2 will correspond to a maximum) Evaluate V''(x) at x = 4: V''(4) = 24(4) - 88 = 8 (Since it's positive value, x = 4 will correspond to a minimum) Therefore, cutting squares of side length 2 inches will result in the largest volume box.
06

Calculate the dimensions and volume of the box

Using x = 2 and the expressions for the dimensions of the box from Step 1: Length: 16 - 2(2) = 12 inches Width: 6 - 2(2) = 2 inches Height: 2 inches Calculate the volume V of the largest box: V = Length 脳 Width 脳 Height V = 12 脳 2 脳 2 V = 48 cubic inches The largest volume box that can be made this way has dimensions of 12x2x2 inches and a volume of 48 cubic inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Maximization
When trying to create the largest possible box from a sheet of cardboard, the focus is on **volume maximization**. The volume of a box is the amount of space it encloses, typically measured in cubic units.
To find this optimal volume, it is crucial to define a function that describes the volume in terms of a variable. In our problem, this variable is the side length, \( x \), of the squares cut from each corner. By changing \( x \), you effectively change the dimensions of the box, and hence its volume. It is from this functional relationship that the largest volume is determined.
  • Volume formula: \( V(x) = (16 - 2x)(6 - 2x)(x) \)
  • Larger volume equates to more space in practical manufacturing.
To find when the volume is biggest, you need to analyze the function and discover its peak.
Critical Points
The key to solving optimization problems lies in finding **critical points**. These are the values of \( x \) that make the derivative of the volume function \( V'(x) \) equal to zero, indicating potential maxima or minima.
To solve for these points, differentiate the volume function. The solutions to \( V'(x) = 0 \) give us the critical points. In this problem, we found the critical points to be \( x = 2 \) and \( x = 4 \).
  • Critical points are where optimization occurs.
  • They represent possible peaks (maximums) or valleys (minimums).
Bear in mind, not all critical points will yield maximum volume, some may be minimum or even points of inflection.
Derivative Applications
Derivatives are mathematical tools used to determine the rate at which a quantity changes. In optimization problems like this, **derivative applications** help us find and verify critical points, which tell us where the maxima or minima occur.
By differentiating the volume function \( V(x) = 4x^3 - 44x^2 + 96x \), we arrived at \( V'(x) = 12x^2 - 88x + 96 \). Setting this equal to zero helps us find our critical points.
Further, the second derivative test can help confirm whether each critical point is a maximum or minimum:
  • Use \( V''(x) = 24x - 88 \) for this test.
  • If \( V''(x) < 0 \), there's a local maximum; if \( V''(x) > 0 \), there's a local minimum.
This method verifies \( x = 2 \) gives the largest volume, while \( x = 4 \) gives a minimum.
Box Dimensions
In box optimization problems, understanding the relationship between **box dimensions** and the cutting pattern is vital. Here, the size of each square cut from the corners (\( x \)) critically affects the final box dimensions.
Once you determine the optimal \( x \), calculate the dimensions: length, width, and height.
In our example:
  • Length: \( 16 - 2x \) becomes 12 inches when \( x = 2 \)
  • Width: \( 6 - 2x \) becomes 2 inches when \( x = 2 \)
  • Height: equals \( x \), or 2 inches
These dimensions directly determine the volume of the optimized box, calculated by multiplying these three values. Thus, greater understanding of dimensions can lead to more efficient material use and better product design.

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