# Find the dimensions that will minimize the cost of the material?

## A cylindrical container that has a capacity of $10 \setminus {\textrm{m}}^{3}$ is to be produced. The top and bottom of the container are to be made of a material that costs $20 per square meter, while the side of that container is to be made of a material costing $15 per square meter. Find the dimensions that will minimize the cost of the material.

Nov 21, 2016

$h = 2.82876 , r = 1.06078$

#### Explanation:

Calling ${c}_{1} = 20$ and ${c}_{2} = 15$ the total cost is

$C = 2 \left(\pi {r}^{2}\right) {c}_{1} + \left(2 \pi r h\right) {c}_{2}$.

Here $r$ is the base radius and $h$ is the side height.

The volume is given by

$V = \pi {r}^{2} h = {V}_{0} = 10$.

Now, the problem can be stated as

${\min}_{h , r} C \left(h , r\right)$ restricted to $V \left(h , r\right) = {V}_{0}$

Using lagrange multipliers it reads

$L \left(h , r , \lambda\right) = C \left(h , r\right) + \lambda \left(V \left(h , r\right) - {V}_{0}\right)$

with stationary points given by

$\nabla L = \vec{0}$ or

{(2 c_2 pi r + pi r^2 lambda= 0), (2 c_2 h pi+ 4 c_1 pi r + 2 h pi r lambda = 0), (h pi r^2 - V_0 =0):}

solving for $h , r , \lambda$ we get

$h = \sqrt[3]{{\left(2 {c}_{1} / {c}_{2}\right)}^{2} {V}_{0} / \pi} , r = \sqrt[3]{{c}_{2} / {c}_{1} {V}_{0} / \left(2 \pi\right)} , \lambda = - 2 \sqrt[3]{\frac{2 \pi {c}_{1} {c}_{2}^{2}}{{V}_{0}}}$ or

$h = 2.82876 , r = 1.06078$ with associated cost $C = 424.214$

We know that is a minimum because

$\left(C \circ V\right) \left(r\right) = \frac{2 \left({c}_{1} \pi {r}^{3} + {c}_{2} {V}_{0}\right)}{r}$ and
$\frac{{d}^{2}}{{\mathrm{dr}}^{2}} \left(C \circ V\right) \left(r\right) = \frac{4 \left({c}_{1} \pi {r}^{3} + {c}_{2} {V}_{0}\right)}{r} ^ 3$

and for the found solution has the value

$\frac{{d}^{2}}{{\mathrm{dr}}^{2}} \left(C \circ V\right) \left(r\right) = 240 \pi > 0$

Nov 21, 2016

For students who have not yet learned calculus of two variables, here is the single variable solution.

#### Explanation:

A (right circular) cylinder has two variables,

height, $h$, and
radius, $r$.

The top and bottom cost $\pi {r}^{2} \left(20\right)$ $. SInce there are two of these, they add $40 \pi {r}^{2} to the cost.

The cost of the side of the cylinder will be $2 \pi r h \left(15\right) = 30 \pi r h$ $The total cost is $40 \pi {r}^{2} + 30 \pi r h$. In order to make this a function of a single variable, we need an equation with both $h$and $r$in it. The volume of a (right circular) cylinder is $V = \pi {r}^{2} h$and we want the volume to be $10 \text{ } {m}^{3}$. $\pi {r}^{2} h = 10$, so $h = \frac{10}{\pi {r}^{2}}$. The cost can now be written as a function of $r$alone. $C \left(r\right) = 40 \pi {r}^{2} + 30 \pi r h = 40 \pi {r}^{2} + 30 \pi r \left(\frac{10}{\pi {r}^{2}}\right)$$= 40 \pi {r}^{2} + 30 \pi r \left(\frac{10}{\pi {r}^{2}}\right)$$= 40 \pi {r}^{2} + \frac{300}{r}$Domain is $r > 0$We want to minimize $C$, so we'll find the derivative, then critical number(s) and then test the critical number(s). $C ' \left(r\right) = 80 \pi r - \frac{300}{r} ^ 2 = \frac{80 \pi {r}^{3} - 300}{r} ^ 2$The only real valued critical number is $r = \sqrt[3]{\frac{15}{4 \pi}}$. Using the first or second derivative test verifies that $C$is minimum at this critical number. Second derivative test: $C ' ' \left(x\right) = 80 \pi + \frac{600}{r} ^ 3$So $C ' ' \left(\sqrt[3]{\frac{15}{4 \pi}}\right) = 80 \pi + \frac{600}{\frac{15}{4 \pi}} > 0$Finally, the question asks for the dimensions, so we have $r = \sqrt[3]{\frac{15}{4 \pi}}$and $h = \frac{10}{\pi {r}^{2}} = \frac{10}{\pi {\left(\sqrt[3]{\frac{15}{4 \pi}}\right)}^{2}}$$= \frac{10}{\pi} {\left(\sqrt[3]{\frac{4 \pi}{15}}\right)}^{2}\$ (Rewrite further as desired.)