# Grace's bakery specializes in giant cupcakes. Each cardboard cupcake container is filled with batter and baked in the oven. The baked cupcake forms a perfect hemispherical on top of the container....

Grace's bakery specializes in giant cupcakes. Each cardboard cupcake container is filled with batter and baked in the oven. The baked cupcake forms a perfect hemispherical on top of the container. The surface area of this is frosted. A clear plastic lid, identical in size and shape to the cupcake container, is then placed on top. These containers are designed such that R1=3/4R2 and h=2R1. Given that 100 cm^3 of batter costs 9cents, 100 cm^2 of frosting costs 27 cents, 100 cm^2 of cardboard costs 1.5 cents and 100 cm^2 of plastic costs 3 cents, complete the tasks below:

A.) Create functions of R1, which return the cost, in dollars, of ingredients, I, and packaging, P. Approximate all coefficients to 4 decimal places.

B.) Find the volume of a baked cupcake if Grace wants 85% of the cost to go toward ingredients. Round to 2 decimal places.

C.) Find the efficiency ratio for the packaging to the baked cupcake, to 2 decimal places.

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embizze | Certified Educator

(1) To find the cost of packaging we need the surface area for the cardboard and plastic (which are the same by symmetry.)

The cup is a right circular frustum. We need the lateral area and the area of the circular base.

The lateral area is given by `LA=pi(R_1+R_2)sqrt((R_1-R_2)^2+h^2) `

We are given that `R_1=3/4R_2 ==> R_2=4/3 R_1 ` and `h=2R_1 ` so the lateral area in terms of R1 is:

`LA=pi(R_1+4/3 R_1)sqrt((R_1 - 4/3 R_1)^2+(2R_1)^2) `

`=pi(7/3 R_1)sqrt(1/9 R_1^2 + 4R_1^2) `

`=pi(7/3 R_1)sqrt(37/9 R_1^2) `

`=pi(7/3R_1)sqrt(37)/3 R_1=(7 pi sqrt(37))/9 R_1^2 `

The area of the circular base is `pi R_1^2 ` so the total area for the cardboard and the plastic is `(7 pi sqrt(37))/9R_1^2+pi R_1^2 ~~ 18.00458R_1^2 `

The cost for the cardboard is 1.5 cents per 100 square cm so the cost for cardboard is `C_(c)=(1.5 )/100  18.00458R_1^2 ~~.27007R_1^2 `

The cost for the plastic is 3 cents per 100 sq cm so the cost for the plastic is `C_p=3/100*18.00458 R_1^2~~.54013 R_1^2 `

The cost for the packaging in cents in terms of the radius of the smaller circular base is `P=.8102 R_1^2 `

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For the cost of the ingredients we need the volume of batter and the area for the frosting.

The volume of batter is the volume of the frustum. (The cake will rise when baked.) The volume is given by `V= (pi h)/3 (R_1^2 + R_1R_2+ R_2)^2 `

Substituting the values for R2 and h we get:

`V=(pi 2R_1)/3 (R_1^2+4/3R_1^2+16/9 R_1^2) `

`=(74pi)/27 R_1^3 ` Since the cost of the batter is 9 cents per 100 cc we have the cost for the batter to be `C_b=9/100 * (74 pi)/27 R_1^3 ~~.77493R_1^3 `

The amount of batter required is the surface area of the hemisphere. The surface area of a hemisphere is `2pi r^2 ` ; we use the radius of the larger circular base or ` R_2=4/3 R_1 ` to get:

`A=2pi(4/3 R_1)^2=(32 pi)/9 R_1^2~~11.17011R_1^2 ` . Since the cost of the frosting is 27 cents per 100 sq cm we have the cost as:

`C_f=27/100 * 11.17011~~3.01592 `

The cost of the ingredients is `I=.77493 R_1^3 + 3.01592 R_1^2 `

So `P=.8102R_1^2, I=.7749 R_1^3 + 3.0159 R_1^2 `

Converting to dollars we get:

`P=.0081R_1^2, I=.0077R_1^3+.0302R_1^2 `

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(2) I=.85T ==> I=.85(I+P)

`.0077R_1^3+.0302R_1^2=.85(.0077R_1^3+ .0383R_1^2) `

`.001155R_1^3=.002355R_1^2 ==> R_1=2.03896 `

So the radius of the smaller base should be 2.04cm.

The volume of the baked cupcake is the volume of the frustum plus the volume of the hemisphere.

`V_t=(pi h)/3 (R_1^2 + R_1R_2 + R+2^2)+ (2pi)/3 R_2^3 `

`=(4.08 pi)/3 (4.1616 + 5.5488 + 7.3984)+(2 pi)/3 (20.1236) `

The volume is approximately 115.25cc