Introducing
Your new presentation assistant.
Refine, enhance, and tailor your content, source relevant images, and edit visuals quicker than ever before.
Trending searches
Optimization Project
3
Optimization Project (maximizing volume)
Calculations
14cm
7cm
Calculations
20.5cm
I chose to maximize the volume of the Great Value™ Naan Crisps. While doing my calculations, I chose to fix the height at 20.5cm because I wanted to ensure that it can still fit in a regular sized shelf. After careful calculations, I have come up with the new dimensions for the maximized box.
Step 1 and 2
The first thing I did was measure the dimensions of the original box to calculate the surface area. The length measured at 14cm, the width at 7cm and the height at 20.5cm. I substituted the dimensions of the length, width and height into the surface area equation of: SA=2lw+2hw+2hw. After substituting in the values, the fixed surface area was 1057cm^2.
The next step was isolating one of two variables to be able to plug it in to the original volume function. I ended up isolating for length, making width the independant variable.
Step 3 and 4
The next step was plugging in the isolated value into the original volume function of V=lwh. This will allow me to calculate the derivative of the volume function.
I needed to find the potential critical points by finding the x-values when the derivative is equal to zero or does not exist. The critical points allow me to identify the maximum and minimum values of my function and would demonstrate the x-intercepts in the domain of the function. It also proves where the graph is increasing and decreasing as it reaches the maximum and minimums.
With the potential critical values, I created a chart to find where the graph increases, decreases and where the local maximum and minimums are located on the graph.
*it is important to note that there can be no negative measurements and that is why any point after zero is crossed out.
Step 5 and 6
After finding the maximum, you can plug the w-value that was found into the volume function and solve for the volume.
The final step is to sub in the w-value into the equation created in step 2 to find the length
*the independant variable is w (width) and the dependant variable is v (volume)
New Design
10.3cm
New design
20.5cm
The new design has the same height as the original box as fixing the height was optimal to ensure that it continues to fit in the shelfs of consumers and grocery stores. The width and length of the new box now has a square base to provide more space for more product.
Description
Description
As previously mentioned, there is a square base on the new box that is made. The reason for this is because at w=10.3cm it reaches the largest length or width that the box can be before the volume starts to decrease again. Any width or length larger than 10.3cm will have a smaller volume than 2175.6cm^3 (maximum volume).
Graphs
Green graph: derivative function
Red graph:
original function
Graphs
vertical asymptote for both graphs at: x=-41/2 or -20.5
Relationships between the two graphs
By observing the blue point that has been plotted you can see that the maximum value on my orginal function graph is at (10.302, 2175.599). In respect to the derivative function graph, the x-intercepts (black point) show the maximum x-value of the original funtion. This is because at the maximum value on the original function the slope is equal to zero. The derivative function displays the slope of the tangent at every point of the original function. Therefore, when the slope of the tangent is equal to zero at x=10.302 on the original graph, the y-value is equal to zero at x=10.302 on the derivative graph.
Relationships between the graph and the new box
Relationship between the new box and the graph
As previously mentioned, the maximum on the graph (blue point) represents the maximum volume (y-value) and the maximum width (x-value). This is related to the volume,the length and width of the new box. In addition, the maximum on the graph also represents the largest length and width that the new box can be otherwise the volume will lower and you would no longer be at your maximum value.
Final Thoughts
Final Decision
I believe that using my new design with a maximized volume will be a better choice for this company.
The new box provides 8% (166.6cm^3) more space for product. This can be used as a marketing strategy. If we announce that we are adding 8% more chips to each box it will entice more customers to want to purchase our product over other competitors. In addition, the shape of the new box is eyecatching and easy to spot in comparison to the other rectangular esqe boxes
There will be a surge in product production, leading to increased profit. The shipping costs will also be reduced as we will be able to ship more boxes out at a time. With a 3.7cm decrease in the length of the box, more boxes will be able to fit in the shipping containers. This can also be reflected in shelf space as more boxes can be packed on each shelf. Therefore, the remaining balance that is not spent on the shipping can go towards production costs.