Finding Area: Solving a Homework Problem

[bondgirl007]

Homework Statement

A printed page is to contain 60cm^2 of printed material with clear margins of 5 cm on each and 3 cm on the top and bottom. Find the minimum total area of the page.

Homework Equations

The Attempt at a Solution

I know how to work these questions but am having difficulty finding the formula for the area. Any help would be appreciated.

 

[dynamicsolo]

Homework Statement

A printed page is to contain 60cm^2 of printed material with clear margins of 5 cm on each and 3 cm on the top and bottom. Find the minimum total area of the page.

Suppose you make the 60 sq.cm. area the basis for setting up the page. Choose variables for its dimensions. What would be the dimensions of the entire page? What function do you want to minimize?

 

[mjsd]

does the printed material have to be in a retangular shape? in fact is the page retangular shaped?

 

[dynamicsolo]

does the printed material have to be in a retangular shape? in fact is the page retangular shaped?

I know it doesn't say so explicitly, but a problem of this type is widely used in calculus texts. The fact that we are told about margins at top and bottom and at either side of the printed area makes it safe to assume that this is a rectangular page. (Moreover, a more general shape would require either more specifications or else multivariate optimization, which is a tad beyond the scope of a typical first calculus course.)

 

[bondgirl007]

The page is rectangle and so is the printed material.

I'm really struggling with this question. :(

 

[mjsd]

it is then a matter of optimizing the border, since the bit in the middle (the printed material) is always at 60cm^2.

 

[dynamicsolo]

The page is rectangle and so is the printed material.

I'm really struggling with this question. :(

How would you express the area of the printed region?

 

[bondgirl007]

How do I do that? What will the formula be for the area?

 

[dynamicsolo]

How do I do that? What will the formula be for the area?

How would you write the area of a rectangle when you don't know the length of its sides yet, but you do know that the area is 60 cm^2?

 

[bondgirl007]

60 = xy?

 

[dynamicsolo]

60 = xy?

Good! Now, given the margins we are instructed to provide around this region, what would the expression for the area of the page be?

 

[bondgirl007]

60=(x+10)(y+6)

Is that it?

 

[dynamicsolo]

60=(x+10)(y+6)

Is that it?

Well, this would be the area we want to minimize, which we could call A. So we'd have (x+10)(y+6) = A. If you multiply that out, what do you get and what do you notice?

 

[bondgirl007]

60 = (x+10)(y+6)
60/(x+10)=y+6
60/(x+10) - 6 = y

60 = (x+10) ( (60/(x+10) -6) + 6
60= 60 - 60(x+10) + 60(x+10)
60=60

There's no variable left now.

 

[dynamicsolo]

60 = (x+10)(y+6)
60/(x+10)=y+6
60/(x+10) - 6 = y

60 = (x+10) ( (60/(x+10) -6) + 6
60= 60 - 60(x+10) + 60(x+10)
60=60

There's no variable left now.

(x+10)(y+6) is equal to an unknown area A, not 60 cm^2. (It's xy that is 60.) You will need to eliminate one of the variables in order to find a minimum for the area function. Which variable would you like to replace?

 

[bondgirl007]

I see.

I'd like to replace y.

 

[dynamicsolo]

I see.

I'd like to replace y.

So y = 60/x . If you put this into the area function, what do you get?

 

[bondgirl007]

A= (x+10)(y+6)
A=(x+10)(60/x + 6)
A = 60+6x + 600/x + 60
A = 120 + 6x + 600/x
dA/dx = 6 - 600/x^2
0 = 6 - 600/x^2
600 = 6x^2
x = 10

and then y = 6.

How does that get me to the answer? The answer for this is 240cm^2.

 

[dynamicsolo]

A= (x+10)(y+6)
A=(x+10)(60/x + 6)
A = 60+6x + 600/x + 60
A = 120 + 6x + 600/x
dA/dx = 6 - 600/x^2
0 = 6 - 600/x^2
600 = 6x^2
x = 10

and then y = 6.

How does that get me to the answer? The answer for this is 240cm^2.

That will be the minimum possible area for the entire page with the margins. If you described the problem correctly, the printed area must be 10 cm. wide and 6 cm. tall (area = 60 cm^2). The top and bottom margins are each 3 cm., so the full page is 3 + 6 + 3 = 12 cm. tall. The side margins are 5 cm. each, so the full page is 5 + 10 + 5 = 20 cm. wide. This gives a total page area of 12 cm. x 20 cm. = 240 cm^2.

That should be the solution to the problem.

 

[bondgirl007]

Thank you soo much!

 

[bondgirl007]

I have another minimum question.

Two towns A and B are 5 km and 7 km, respectively, from a railroad line. The points C and D nearest to A and B on the line are 6 km apart. Where should a station be located to minimize the length of a new road from A to S to B?

I have no clue about the formulas or anything in this one.

 

[dynamicsolo]

Two towns A and B are 5 km and 7 km, respectively, from a railroad line. The points C and D nearest to A and B on the line are 6 km apart. Where should a station be located to minimize the length of a new road from A to S to B?

This is one where it's a good idea to draw a picture. With a line to represent the railroad, pick two points on it and call the distance separating them 6 km. Draw a perpendicular line from one of the points out to a point A and make that distance 5 km. From the other point on the railroad, draw a second perpendicular line out to a point B and make that distance 7 km. For the purposes of the problem, it doesn't matter if A and B are on the same side of the rail-line.

We are looking for a point S (for the new rail station) between the two points on the rail-line so that the length of a line drawn from A to S plus the length of a line drawn from S to B, which represents the road from A to S to B, is as short as possible.

 

[bondgirl007]

Thanks for the reply. I've drawn that picture.

What formula should I use though?
A+S + B+S = ASB?

 

[dynamicsolo]

Thanks for the reply. I've drawn that picture.

What formula should I use though?
A+S + B+S = ASB?

You want to use the distance formula to find the straight-line distances from A to S and from B to S, since we are presumably building straight roads to the station from each town. The total length of road needed will be the sum of those distances.

Where does you put the unknown that is going to be the variable that you minimize this distance with respect to?

 

[bondgirl007]

Will the formula be:
d = AS + BS?

What variable am I supposed to isolate?

Sorry but I'm really confused.

 

[bondgirl007]

The answer for this is 21/6 km from D.

 

[dynamicsolo]

Will the formula be:
d = AS + BS?

Yes.

What variable am I supposed to isolate?

You'll need to identify the distance of the station from one or the other of the two points on the rail line from which you drew the perpendicular lines out to the towns. The road to each town will have formed a right triangle using each set of the lines you've drawn. The length of each road is the length of the hypotenuse of that right triangle.

You can now make a total distance function which is the total length of road to be built, using the distance I described as the variable.

 

[bondgirl007]

This is what my picture looks like.

http://img223.imageshack.us/my.php?image=24392478ko2.jpg

Do I use trig to find the distances?

 

[bondgirl007]

I have the two right triangles but not sure how to find the distance from C to S or the distance from D to S. The whole thing is 6 km so will it just be 3 km and then do I find the hypotenuse?

 

[dynamicsolo]

I have the two right triangles but not sure how to find the distance from C to S or the distance from D to S. The whole thing is 6 km so will it just be 3 km and then do I find the hypotenuse?

This is where you pick your variable. Call either the distance from C to S or from D to S, x. What will the other distance be?

How would you now find the distance from A to S and from B to S? (Triangles ACS and BDS are right triangles.)

 

[bondgirl007]

I'd find it with Pythagoras.

So will it be:

(5)^2 + (3)^2 = AS^2
AS = 5.83

 

[dynamicsolo]

I'd find it with Pythagoras.

So will it be:

(5)^2 + (3)^2 = AS^2
AS = 5.83

The station isn't necessarily going to be at the midpoint, so you can't just use a '3' for that one leg CS (in fact, I'll bet money that leg won't be 3). Let's call that leg x. Then we have

5^2 + x^2 = AS^2 .

What would you write for BS^2?

 

[bondgirl007]

For BS^2, I'd write:

7^2 + (6-x)^2 = BS^2.

 

[bondgirl007]

After these formulas, I solved for AS and BS and this is what I got:

AS = 5+x
BS = 13-x

And then I substituted them into the distance formula:

d = AS + BS
d= 5+x + 13-x
da/dx = 0

I'm really confused now.

 

[dynamicsolo]

For BS^2, I'd write:

7^2 + (6-x)^2 = BS^2.

Good! Now, in principle, you would take the square root of each of these expressions to get AS and BS individually. The sum of these two lengths will be the total length of the road. You would take a derivative of that sum to work out the critical point of this function. That value of x would tell you where to put the station to minimize the length of road.

(I take back part of what I typed. You are only asked to find where to put the station, so you just need to find x and 6-x .)