Little Hershey Kiss

February 23, 2001

Hershey Food Corporation

19 E. Chocolate Ave.

Hershey, PA 17033

Dear Chocolate Lovers:

This is probably the worst thing I have ever heard of with losing the recipe to the famous Little Hershey Kiss. I understand having chocolate lovers everywhere and not being able to produce the Kiss could be catastrophic to the human race. You came to the right people for help with this problem since we have helped other large corporations with manufacturing problems. You stated in your letter that you know what ingredients you need for the recipe since it doesn’t vary from your other lines of candy. You also identified that you need to know the quantity of chocolate used to make the perfect little Hershey Kiss. Being a Kiss lover ourselves we just happen to have some on hand. We will use a piece of candy along with a little calculus to determine the amount of chocolate needed. In the letter you indicated the need to know the size of the piece of aluminum foil to wrap the Kiss. We can also apply calculus to find this for you.

We will first cut a Kiss in half so we can figure the exact size of the Kiss. We placed it on a piece of graph paper scaled in millimeters and traced around only one side of it as shown below.

Using this graph we will pick twelve data points with respect to height and diameter on the outline of the HersheyKiss which are shown below.

Height / Diameter
0 / 11
3 / 12
5 / 12
7 / 11
8 / 10
10 / 9
11 / 8
13 / 6
14 / 5
15 / 4
20 / 2
22 / 0

Using Excel we developed a clearer graph:

Using our TI 89 to graph the data from the table we were able to get an equation of the graph:

F(x) = 0.003313435981271x^3 - 0.12720470645299x^2 + 0.71886532976505x + 11.021944615485

Using this equation we can find the volume of the Hershey Kiss by integrating the equation from zero to twenty-two. When doing this we use the equation:

V = (PI*r^2*h)

We substitute the r (radius) with the equation F(x) and the h (height) with dx.

V = (PI*F(x)^2*dx)

Using the TI 89 the given results are 4798.41 mm^3 which is the amount of chocolate in one Hershey kiss. You also stated in your letter that you needed to have the results measured in ml to use with your equipment at the chocolate factory. To do this first we convert mm^3 to cm^3 by dividing 4798.41 by 1000 to equal 4.79841 cm^3. One cm^3 is equal to one ml so the amount is 4.798 ml of chocolate.

You are interested in a second method to determine in the amount of chocolate in a Kiss. This method did not involve Calculus. Our first idea was to melt the Kiss over boiling water. This was unsuccessful mostly because we didn’t have the right equipment. The only thing we ended up with was a mess in the kitchen!

Using more Kisses we successfully used water displacement to get a volume measurement. We removed the wrappers from 4 Kisses and put them in a beaker holding 30 cc of water. We noted the water level rose to 46.5 cc. Since the volume increased by 16.5 cc we divided this number by 4 to get the average volume for 1 Kiss = 4.125 ml ( 1cc = 1 ml).

To find the amount of foil used to wrap the Hershey Kiss we need to find the surface area of the Kiss. Doing this we integrate from zero to twenty-two again by using the equation:

SA = 2PI F(x)*(1+(F’(x))^2)^1/2 dx

To clarify the expression in the equation:

SA = Surface Area

F(x) = Our original equation

F’(x) = The derivative of the original equation

The derivative of our original function is as follows:

F’(x) = (0.00994x^2 – 0.254409x + .718865)

Substitute these into our equation and with use of the TI 89 we get 1164.21 mm^2.

To get the total surface area we need to add the dimension of the bottom of the Kiss found from the radius with using the equation for the area of a circle:

Area of a circle = Pir^2

We found our radius to be 11mm and plugging this into the equation we get 121PI. When adding together the results we find the amount of aluminum foil to be 1544.34 mm^2.

To sum up our Calculus findings each Hershey Kiss contains 4.798 ml of chocolate. The water displacement method average was 4.125 mlof chocolate. The amount of foil used to wrap each Kiss is 1544.34 mm^2. I’m hoping since you now have these figures that production can resume immediately so that store shelves will not be “Kiss-less”! Chocolate lovers around the world can rejoice!

Thank you for asking our team to assist with your dilemma.

Sincerely,

Dave Thomas

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