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1. A torus donut is formed by revolving the circle (x-a)^2 + y^2 = b^2 around the y-axis. (Here b is the radius of the circle and a is the distance from the center of the hole to the center of the circle.) At present the donuts are made with a=5cm and b=3cm. With the volume fixed at 90pi^2, find the dimensions a and b of the donut which minimize the surface area of the donut. For this problem, you will have to determine the range of allowable a and b.

2. Is there a maximum surface area for the given volume?

3. Write a report to the CEO summarizing your recommendations (including the percent savings or percent extra cost).

1. To find the volume of that torus doughnut, look in http://whistleralley.com/torus/torus.htm

It comes down to the volume being (πrē)(2πR). Looking at the sheet myself gives us r as b and R as a.

2. If we been given a=5cm and b=3cm, the surface area is fixed.

If we wanted to maximize the surface area, we would have to be allowed to vary these values.

In doing so, it could be seen that as a is increased, b would need to be decreased.

If could also be seen that increasing a indefinitely would increase the surface area,

but being the doughnut is made from bread dough, increasing a by very much would cause the doughnut to quickly fall apart.

3. I'm not sure exactly what would be desired here. What I would recommed would be leaving the dimensions as they are and adding something to give a little waviness to the surface, like putting a few ridges on the inside of the doughnut maker. In this way, the surface area would be increased at a minimal charge. The ridges could be made by altering the inside of the doughnut press, but then that might add to cost when considering the increased difficulty in cleaning.

Calculus

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