Help on Precalculus - Maximums and Minimums?

Question

Suppose you have to use exactly 200m of fencing to make either one square enclosure or two separate square enclosures of any sizes you wish. What plan gives you the least area? the greatest area?

and

In a rectangular piece of cardboard with perimeter 30 in., two parallel and equally spaced creases are made. The cardboard is then folded to make a prism with open ends that are equilateral triangles.

a. Show that the volume of the prism is V(x) = (sqrt3/4x^2)(15-3x)

b. What is the domain of V?

c. Find the approximate value of x that maximizes the volume. Then give the approximate maximum volume.

In all honestly, I have no clue where to start on either problems. Can someone please help me?

Answer 1

First problem:

No matter what, each enclosure has to be a square.  First, the only square you can obtain with 200m fencing is 50m x 50m.  This area equals to 2500m.

For the enclosures where you have two squares, I first tested 25m x 25m for each.  That equals to 1250m total.  Then I changed the dimensions to two squares of 40m x 40m and 10m x 10m.  This equals 1700m both.  This shows me that the minimum area is 1250 and will never go below that.

However, using logic, neither of the two squares will ever be 50m x 50m, because then there will only be one square.  So then the maximum area you have is 2500m with only one square.  (To test this I used 1m x 1m and 48.5m x 48.5m.  The total here was 2353.25m.)

Second problem:

Are you sure you copied the Volume correctly in part a?  Because I am getting close, but different answers for Volume than what you have.  Part a is telling you the answer and wants you to show the work, but I am not getting that answer.  Unless someone else can do the second problem.

Answer 2

Problem 1 already answered.

Problem 2.

(a)

L = (30 - 6x)/2 = 15 - 3x

V = (1/2)(volume of enclosing box)

Volume of box = (x)[x*SQRT(3)/2](15 - 3x)

V = (1/2){ (x)[x*SQRT(3)/2](15 - 3x)}

V = [SQRT(3)/4](x^2)(15 - 3x)

(b) domain is 0 to 5 both of which will give a volume of 0.

(c) you can get an approximate value by graphing it. This shows the max is around x = 3.33 and that the volume is 48.11

You can also do some calculations if you want the more difficult way of estimating, Something like Excel does it much more quickly. But another way is to find values of x that bracket the max and then keep dividing the region in two until you reach some value that you decide is close enough.