Question

Find the radius and height of a cylindrical soda can with a volume of 256cm^3 that minimize the surface area.

Answer 1

Answer:

A) Radius: 3.44 cm.

Height: 6.88 cm.

B) Radius: 2.73 cm.

Height: 10.92 cm.

Step-by-step explanation:

We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.

a) We can express the volume of the soda can as:

$V=\pi r^2h=256$

This is the constraint.

The function we want to minimize is the surface, and it can be expressed as:

$S=2\pi rh+2\pi r^2$

To solve this, we can express h in function of r:

$V=\pi r^2h=256\\\\h=\frac{256}{\pi r^2}$

And replace it in the surface equation

$S=2\pi rh+2\pi r^2=2\pi r(\frac{256}{\pi r^2})+2\pi r^2=\frac{512}{r} +2\pi r^2$

To optimize the function, we derive and equal to zero

$\frac{dS}{dr}=512*(-1)*r^{-2}+4\pi r=0\\\\\frac{-512}{r^2}+4\pi r=0\\\\r^3=\frac{512}{4\pi} \\\\r=\sqrt[3]{\frac{512}{4\pi} } =\sqrt[3]{40.74 }=3.44$

The radius that minimizes the surface is r=3.44 cm.

The height is then

$h=\frac{256}{\pi r^2}=\frac{256}{\pi (3.44)^2}=6.88$

The height that minimizes the surface is h=6.88 cm.

b) The new equation for the real surface is:

$S=2\pi rh+2*(2\pi r^2)=2\pi rh+4\pi r^2$

We derive and equal to zero

$\frac{dS}{dr}=512*(-1)*r^{-2}+8\pi r=0\\\\\frac{-512}{r^2}+8\pi r=0\\\\r^3=\frac{512}{8\pi} \\\\r=\sqrt[3]{\frac{512}{8\pi}}=\sqrt[3]{20.37}=2.73$

The radius that minimizes the real surface is r=2.73 cm.

The height is then

$h=\frac{256}{\pi r^2}=\frac{256}{\pi (2.73)^2}=10.92$

The height that minimizes the real surface is h=10.92 cm.