Calculus Max Min Surface Problems

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On the Surface
Max Min Surface Problems

In the surface type problem, we usually need to find the minimum surface area

needed to enclose a certain volume of some stuff.

The key to this problem is the shape of the container.

Often, the container is a cylinder (like a can).

The equations for this shape container are a little more complex

than those for a box.

Example:

What is the least surface area can (cylinder)

that will enclose 15 cubic inches of stuff.

The can DOES have a lid.

To solve this,

we first need the equation for the surface area and volume of a cylinder.

So we have two unknown variables, the radius (r) and the height (h).

We need to solve for one in terms of the other,

so that we can do calculus on a Surface Area equation

with just one independent variable.

We know the total volume (15 in³), so we can use the volume equation to do that...

Volume (V) = pr²h

15 = pr²h

h =	15

	pr²

Substituting this into the surface area equation for h, we get ...

A = 2pr(	15	) + 2pr²

	pr²

Simplifying this:

A =	30	+ 2pr²

	r

Taking the derivative of this,

Remember, r is the variable and p is just a number

A = 30r^-1 + 2pr²

A' = -30r^-2 + 4pr

Find where the slope is zero ...

0 = -30r^-2 + 4pr

30r^-2 + 4pr

Multiply both sides by r ²

30 + 4pr³

2.387 = r³

= r

1.336 » r

When we solved for h, we found that

h =	15

	pr²

h »	15

	p(1.336)²

h » 2.675

So the surface area of the can is:

A	= 2prh	+ 2pr²
A	= 2p(1.336)(2.675)	+ 2p(1.336)²
A	= 22.455	+ 11.215
A	= 33.670 square inches

To check to see if this is a minimum or maximum area, find the second derivative

find the second derivative of the Area equation and check for concavity.

POP QUIZ:

We are looking for a MINIMUM area,

do we want the second derivative value to be positive or negative?

A' = -30r^-2 + 4pr

A'' = 60r^-3 + 4p

Which we can write as ...

A'' =	60	+ 4p

	r³

Looking at this, you can probably see that the only place that A'' can be negative

is where r is negative.

And a negative radius has no meaning in this problem.

The radius value at the A' = 0 point was 1.336,

so plug that into the second derivative equation ...

A'' =	60	+ 4p

	(1.336)³

A'' = 44.910 + 12.566

A'' = 57.476

57.476 is a positive number so the area is a minimum.

And that area was 33.670 square inches. THAT IS OUR ANSWER.

It is WAY common in these problems to work everything out,

but not specify what your answer is.

Some teachers will cut you slack on that one,

but most will hit you for at least a few points.

Let's look at those dimensions again.

The radius we found was 1.336, that makes the diameter 2.672

The height we found was 2.675

Except for rounding error, that is the same value.

Interesting eh?

The smallest surface area of any shape for a given volume is a sphere.

The smallest surface area of a box type shape for a given volume is a cube.

A can with the same diameter and height

is the cylinder equivalent of a sphere or cube.

A variation on this problem

is where the material used for the different parts of the container

cost different amounts of money per square area unit.

The problem is to spend the least amount of money on the container.

This is a nearly identical problem to the one we just did.

If we changed the problem we did

and said that the material for the top and bottom of the cylinder

cost $3 per square inch and the material for the sides cost $2 per square inch.

Now to find the least amount of money we would spend building the container,

the area equation:

A = 2prh + 2pr²

Would become the cost equation:

$ = ($2) x 2prh + ($3) x 2pr²

And we would work the problem the exact same way.

1) Substitute for h in the volume equation

2) Find the derivative of the $ equation, that is, $'

3) Set $' equal to zero and solve for r

4) Use the volume equation to find h

5) Find the second derivative of $, that is, $'' and check for concavity.

Hmmm, sounds like a good problem to use in homework, eh?

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