Finding Dimensions and Volumes of Rectangles

[spyder1125]

I have 3 problems I'm having difficulty with in figuring out which correct equation to use to solve them. One deals with Minimizing Packaging Costs, the other Parcel Post Regulations, and finally Book Design.

Here are the problems, the attempts that I did follow them. I guess I just need a step in the right direction on starting out the problems, then I can work through the math. Thank you in advance!

Minimizing Packaging Costs:A rectangular box is to have a square base, and volume of 20 ft. If material for base is $.30/square foot, $.10/sq ft for sides, and $.20/sq ft for top, determine the dimensions of the box that can be built at minimum cost.

Does that mean we use the v=lwh? But then there are 3 different numbers. I used the s=2x^2+4(xy) but then I got stuck doing that....so I'm not sure which one to do for that, it doesn't appear that the p=2x+y and c=12x + 3y

Parcel Post Regulations:A parcel sent may have a combined length and girth of no more than 108 in. Find the dimensions or a rectangular package that has a square cross section and the largest volume that may be sent through the mail. What is the volume of the package?
Hint: The length plus the girth is 4x + h

I did the s=2x^2 + 4xy formula, solved it out and got x=3.7798...but then I dont even know if that was the right formula to use for that one. I'm assuming that since they want the volume, then we have to do the v=lwh and v=x^2y =108???

Book Design:A book designer decided the pages should have 1-in. margins at top and bottom and 1/2-in. margins on sides. Each page should have area of 50 in. squared. Determine the page dimensions that will result in maximum printed area on that page.

i used the v=lwh, v=x(2x-1)(2y-2) ???? i doubled the 1/2" margin for the x value and subtract that away, and did the same with the y dimension of the book, double the 1" margin and subtracted from the y dimension. So i'm left with 4x^2y - 4x^2 - 2xy + 2x.....so now i'm being thrown off with that y in the equation, which is something we haven't done before in class, which leads me to believe that the formula i'm using is totally wrong

 

[mark07]

Minimizing Packaging Costs:A rectangular box is to have a square base, and volume of 20 ft. If material for base is $.30/square foot, $.10/sq ft for sides, and $.20/sq ft for top, determine the dimensions of the box that can be built at minimum cost.

Let x be the side of the square base and h be the height of the box.

You want to minimize the cost. Let's write the cost function:

\(\displaystyle \L C(x,h) = 0.3x^2 + 0.1 (4xh) + 0.2 x^2\)

They also gave the volume so you can express h in terms of x (or vice versa) and write your cost function using only one variable.

\(\displaystyle \L \text{Volume } = x^2 h = 20\)

\(\displaystyle \L h = \frac{20}{x^2}\)

Then the cost function becomes

\(\displaystyle \L C(x) = 0.5x^2 + \frac{8}{x}\)

and you can minimize it.

 

[spyder1125]

thanks that helps a lot with the first one....can you lead me in the right direction with the other two as well. I'm stuck on those also.

 

[stapel]

The others are related and somewhat similar to the one for which you've been provided the set-up. How far have you gotten on them? Where are you stuck?

Please be specific. Thank you.

Eliz.

 

[spyder1125]

For the Parcel Post Regulations problem, this is what I've done so far:

v=lwh
v=x^2y=108
y=108/x^2

s= 2x^2 + 4(xy)
s= 2x^2 + 4x (108 / x^2)
s= 2x^2 + (432x / x^2)
s= 2x^2 + 432x^-1
s'= 4x - 432x^-2
s''= 4 + 864x^-3

let the 2nd derivative equal to zero

4x - 432x^-2 = 0
4x = 432x^-2
4x = 432 / x^2
4x^3 = 432 / 2
4x^3 / 4 = 216 / 4
x^3 = 54
x = 3.7798

For the Book Design problem, this is what I have so far, and I'm stuck:

v=lwh
v= x(2x-1)(2y-2)
v = 4xy - 4x - 2y + 2
v = 4x^2y - 4x^2 - 2xy + 2x

I've never had the equation with both the x and y variable in it before like this, which leads me to beleive that i even started the problem wrong.

By the way, how is it that the mathematics appears visually better on mark07's post?

Thanks again for you help!

 

[mark07]

2) For the Parcel Post Regulations problem, this is what I've done so far:

v=lwh
v=x^2y=108
y=108/x^2

Volume is what you want to maximize. 108 is combined length and girth, i.e. 4x+h=108.

V=x^2 h

To write that in terms of x only, solve 4x+h=108 for h, and put it in V. Then do the calculus.

3) For the Book Design problem, this is what I have so far, and I'm stuck:

v=lwh
v= x(2x-1)(2y-2)
v = 4xy - 4x - 2y + 2
v = 4x^2y - 4x^2 - 2xy + 2x

Problem is talking about a page. Why are you setting up a volume?

Let x be the width of the page and y be the height.

The area of the full page is 50, so you got xy=50. This is the equation you use to solve y for x later.

Your objective is to maximize the printed area. Write a function for the area using x,y, and the margin restrictions. Then using the above equation, express your function using only x. And do the calculus to find its maximum.

Hope this helps.

By the way, how is it that the mathematics appears visually better on mark07's post?

I use TeX codes. If you browse the Forum Help on top of the page, you can see some links. For beginners, I think TeXaide is an excellent free program. You can prepare your equations (without knowing any TeX) in the program and it gives you the TeX code that you can paste in your message between [ tex ] [ /tex ] (without spaces of course).

 

[spyder1125]

Your objective is to maximize the printed area. Write a function for the area using x,y, and the margin restrictions. Then using the above equation, express your function using only x. And do the calculus to find its maximum.

so for the book design function, would that be something like:

A = (2x - 1)(2y - 2)
A = 4xy - 4x - 2y + 2

Is that right? How would I go about continuing in solving for x?[/quote]

 

[mark07]

Spyder,

Get a piece of paper. Call the width x, height y. You are given that area of your paper is 50, i.e. xy=50, then y=50/x.

Top and bottom margins are 1 each. Side margins are 1/2 each. So the area of the printed part is

A = (x - 1/2 - 1/2)(y - 1 - 1)

isn't it? Substitute y=50/x in there.

\(\displaystyle \L A(x) = (x-1)\left( \frac{50}{x} - 2 \right)
= 50 - \frac{50}{x} -2x +2\)

Find A'(x), solve A'(x)=0 for x.