A high-class chocolate box adds a strip of width x down across the front of the cookie box. Find the new volume V(x) and the x that maximizes it. Extra credit: Show that V_max is r... A high-class chocolate box adds a strip of width x down across the front of the cookie box. Find the new volume V(x) and the x that maximizes it. Extra credit: Show that V_max is reduced by more than 20%.
Understand the Problem
The question is asking to find the new volume function V(x) of a cookie box after adding a strip of width x, and also to determine the value of x that maximizes this volume. Additionally, there's a request for extra credit to show that the maximum volume is reduced by more than 20%.
Answer
New volume function: \( V(x) = (L - x) \times W \times H \); maximizer \( x = 0 \); reduction exceeds 20% for significant \( x \).
Answer for screen readers
The new volume function is ( V(x) = (L - x) \times W \times H ). The value of ( x ) that maximizes the volume is ( x = 0 ), meaning the maximum volume ( V_{\text{max}} = V_0 ). The reduction in volume exceeds 20% for sufficiently large ( x ).
Steps to Solve
- Define the Original Volume Function
Assume the original dimensions of the cookie box are length ( L ), width ( W ), and height ( H ).
The original volume ( V_0 ) of the box is given by: $$ V_0 = L \times W \times H $$
- Determine the New Volume Function with Width x
When a strip of width ( x ) is added across the front, the new dimensions become:
- New length = ( L - x )
- Original width remains ( W )
- Original height remains ( H )
The new volume function ( V(x) ) will be: $$ V(x) = (L - x) \times W \times H $$
- Maximize the Volume Function
To find the value of ( x ) that maximizes ( V(x) ), we need to take the derivative ( V'(x) ) and set it to zero:
$$ V'(x) = -W \times H $$
Since the derivative is constant and negative, it implies that ( V(x) ) is a decreasing function. Thus, the maximum volume occurs at ( x = 0 ).
- Calculate the Maximum Volume
Substituting ( x = 0 ) back into the volume function yields: $$ V_{\text{max}} = V(0) = L \times W \times H = V_0 $$
- Determine the Volume Reduction
To show that the maximum volume is reduced by more than 20%, we compare V_max and V_min (when ( x ) is maximum):
$$ V_{\text{min}} = (L - x) \times W \times H $$
If ( x ) is significant enough (especially if ( x > 0.2L )), then: $$ |V_{\text{max}} - V_{\text{min}}| > 0.2 \times V_{\text{max}} $$
- Extra Credit Verification
Calculate the percentage reduction: $$ \text{Percentage Reduction} = \frac{V_{\text{max}} - V_{\text{min}}}{V_{\text{max}}} \times 100 $$
Plugging in the values will demonstrate the reduction is over 20%, provided ( x ) is large enough.
The new volume function is ( V(x) = (L - x) \times W \times H ). The value of ( x ) that maximizes the volume is ( x = 0 ), meaning the maximum volume ( V_{\text{max}} = V_0 ). The reduction in volume exceeds 20% for sufficiently large ( x ).
More Information
The original volume is maintained only when ( x = 0 ). The goal of maximizing volume while accounting for the strip width results in reduced volume when ( x ) increases.
Tips
- Ignoring dimensions: Ensure all dimensions are accounted for when calculating the new volume.
- Derivative errors: Mistakes in differentiation can lead to wrong values for ( x ).
- Percentage calculation: Be careful with the percentage reduction calculations; improper setup can lead to incorrect conclusions.
