Questions: A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is 10 times as great for the hemisphere as it is for the cylindrical sidewall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction. If r is the radius of the hemisphere, h the height of the cylinder, and V the volume, then r= and h=. (Type exact answers, using r as needed.)

Solution Steps

To solve this problem, we need to minimize the cost of construction given the constraints. The total volume of the silo is the sum of the volume of the cylinder and the hemisphere. The cost is determined by the surface area, with the hemisphere being more expensive. We will express the cost in terms of the radius and height, and use calculus to find the dimensions that minimize this cost.

1. Express the volume constraint: $V = \pi r^2 h + \frac{2}{3} \pi r^3$.
2. Express the cost function: Cost = $2\pi r h + 10(2\pi r^2)$.
3. Use the volume constraint to express $h$ in terms of $r$.
4. Substitute $h$ into the cost function to get a function of $r$ only.
5. Differentiate the cost function with respect to $r$ and find the critical points.
6. Determine the dimensions $r$ and $h$ that minimize the cost.

The volume $V$ of the silo, which consists of a cylindrical part and a hemispherical part, is given by the equation: $$V = \pi r^2 h + \frac{2}{3} \pi r^3$$ Rearranging this equation to express $h$ in terms of $r$ and $V$, we have: $$h = \frac{V}{\pi r^2} - \frac{2}{3} r$$

The cost $C$ of constructing the silo is determined by the surface area, with the cost of the hemisphere being 10 times that of the cylindrical sidewall. The cost function can be expressed as: $$C = 2\pi r h + 10(2\pi r^2)$$ Substituting the expression for $h$ into the cost function gives: $$C = 20\pi r^2 + 2\pi r \left(\frac{V}{\pi r^2} - \frac{2}{3} r\right)$$ This simplifies to: $$C = 20\pi r^2 + \frac{2V}{r} - \frac{4\pi}{3} r^2$$

To minimize the cost, we differentiate $C$ with respect to $r$ and set the derivative equal to zero: $$\frac{dC}{dr} = 2\pi r \left(-\frac{V}{r^3} - \frac{4}{3}\right) + 40\pi r + \frac{-2V}{r^2} = 0$$ Solving this equation yields the critical points: $$r = 0.2574 V^{1/3}$$ The other critical points are complex and not relevant for this physical problem.

Using the critical point $r = 0.2574 V^{1/3}$, we can substitute back into the expression for $h$: $$h = \frac{V}{\pi (0.2574 V^{1/3})^2} - \frac{2}{3} (0.2574 V^{1/3})$$ Calculating this will give us the height corresponding to the optimal radius.

Final Answer