1. ## Optimization

Hey there. I basically understand the whole method behind these kinds of problems but on one question in particular I am stuck as to how to set up the equation.

There is a solid formed by joining a hemisphere (half a sphere) to each end of a right circular cylinder. The Volume must be 3000 feet. The hemispherical ends cost twice as much as the surface area of the sides. i have to find the dimensions that minimize the cost.

For V i set up the equation v= 4/3(pi)(r^2) + (pi)(r^2)(h)

Is this right? If it is, what variable do i set the equation equal to, h or r?

Finally, can someone explain how to come up with the cost equation?
After that I can take it from there, hopefully. Thanks

2. Originally Posted by painterchica16
Hey there. I basically understand the whole method behind these kinds of problems but on one question in particular I am stuck as to how to set up the equation.

There is a solid formed by joining a hemisphere (half a sphere) to each end of a right circular cylinder. The Volume must be 3000 cubic feet. The hemispherical ends cost twice as much as the surface area of the sides. i have to find the dimensions that minimize the cost.

For V i set up the equation v= 4/3(pi)(r^3) + (pi)(r^2)(h)

Is this right? If it is, what variable do i set the equation equal to, h or r?

Finally, can someone explain how to come up with the cost equation?
After that I can take it from there, hopefully. Thanks
Your considerations are correct so far.

According to the problem you know:

$\displaystyle \dfrac43 \pi r^3 + \pi r^2 h = 3000$ ...... [1]

Now calculate the surface area:

$\displaystyle a=\underbrace{\underbrace{2 \pi r h}_{cylindrical} + \underbrace{4 \pi r^2}_{spherical}}_{parts}$ ...... [2]

Costs:

$\displaystyle c = 2\pi r h + 2 \cdot 4 \pi r^2$ ...... [3]

Calculate h from [1] and plug in this term into [3]. You'll get a function of the costs wrt r:

$\displaystyle c(r) = 2\pi r \left( \dfrac{3000}{\pi r^2} - \dfrac43r \right) + 2 \cdot 4 \pi r^2$

Now determine the minimum of c(r).
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