Question

Suppose a cylinder that holds 3 L of liquid must be created. Determine the radius and height of the cylinder that will minimize the amount of material used in its construction.(Note: 1 L=1,000 cm^3)

Answer 1

The radius and the height of the dimension that will minimize the amount of material used in the construction are $\mathbf{r = \sqrt[3]{\dfrac{1500}{\pi}} }$ and $\mathbf{h = \dfrac{3000}{\pi ({\dfrac{1500}{\pi}} )^{2/3}}}$ respectively.

How to find the dimension that minimizes a cylinder?

The dimension that minimizes the surface area of a cylinder can be determined by:

• Drawing the picture of the problem,
• Write down & identify optimization as well as the constraint equations;
• Use the derivative of the optimization equation to find the dimensions.

Given that:

• 1 L = 1000 cm³
• 3 L = 3000 cm³

The area of the cylinder = (2 πr)h + 2(πr²)

• A = 2πrh + 2πr²

The volume of the cylinder

• V = πr²h

Let's identify the constraint equation and Optimization equation:

To minimize the surface area of the can, we have:

• Area equation = Optimization equation

The constraint equation is the equation that limits us:

• Volume equation = constraint equation

So, Let's solve for h in our volume equation, we have:

3000 = πr² h

h = 3000/πr²

Now, from the Area equation

$\mathbf{A = 2 \pi r(\dfrac{3000}{\pi r^2}) + 2\pi r^2}$

Taking the derivate and setting it to zero, we have;

Derivative:

$\mathbf{A = \dfrac{6000}{ r}+ 2\pi r^2}$

$\mathbf{A = 6000 r^{-1} + 2\pi r^2}$

$\mathbf{A' = -6000 r^{-2} + 4\pi r}$

$\mathbf{A' = 4\pi r-\dfrac{6000}{ r^{2} }}$

Setting it to zero, we  have:

$\mathbf{0=\dfrac{ 4 \pi r^3 - 6000}{r^2}}$

Factor out 4

0 = 4(πr³ - 1500)

1500 = πr³

r³ = 1500/π

$\mathbf{r = \sqrt[3]{\dfrac{1500}{\pi}} }$

The above is the radius that minimizes the surface area of the cylinder;

From $\mathbf{h = \dfrac{3000}{\pi r^2}}$

$\mathbf{h = \dfrac{3000}{\pi ( \sqrt[3]{\dfrac{1500}{\pi}} )^2}}$

$\mathbf{h = \dfrac{3000}{\pi ({\dfrac{1500}{\pi}} )^{2/3}}}$

Thus, the radius and height that minimize the amount of material to be used in its construction are $\mathbf{r = \sqrt[3]{\dfrac{1500}{\pi}} }$ and $\mathbf{h = \dfrac{3000}{\pi ({\dfrac{1500}{\pi}} )^{2/3}}}$ respectively.

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