# The Damned Open-Top Box

1. Aug 22, 2007

### omg precal

1. The problem statement, all variables and given/known data

An open-top box is to be made so that its width is 4 ft and its volume is 40 ft^3. The base of the box costs $4/ft^2 and the sides cost$2/ft^2.

a. Express the cost of the box as a function of its length l and height h.
b. Find a relationship between l and h.
c. Express the cost as a function of h only.
d. Give the domain of the cost function.
e. Use a graphing calculator or computer to approximate the dimensions of the box having least cost.

2. Relevant equations

NONE

3. The attempt at a solution

a. I got that. f(l,h) = 8lh + 16l.
b. I got that. lh = 10 or l = 10/h.
c. What the living hell????????
d. I have 1000% ABSOLUTELY no idea whatsoever.
e. This bugs me. The lower the length is, the lower the total cost. I answered 0.000...01 (infinite). This way the cost is lowest. But it seems so illogical!

This is my Precalculus summer assignement. I can't get these hard-as-hell problems wrong. I need this urgently and ASAP!!! Thanks in advance.

2. Aug 22, 2007

### AlephZero

In (a) you seem to be assuming the box has a base and two sides. (Or you assumed the box has two sides and two ends, but the ends cost nothing).

Part (e) will make more sense if you start again and assume the box has a base and four sides.

For (c): the answer to (b) is an equation connecting l and h. You can use that to eliminate l from the answer to (a).

3. Aug 22, 2007

### omg precal

Umm, no I'm not.

l * h gives me a side. each side costs 2 dollars/sq ft. there are 4 sides. 4*2*l*h = 8lh.

l * w gives me the base. w = 4. 4 * l gives me the base. The base costs 4 dollars/sq ft. 4*4*l = 16l

the total cost is 8lh + 16l.

and how can i remove l if it is necessary for figuring out the cost?

e. say that the length is 1, the width is 4, and the height is 10. the volume is 40. the base would be 4 sq ft, costing 16 dollars. each of the sides would be 10 sq ft, costing 20 dollars each. that gives me a total of 96 dollars.

now say that the length is 0.1 ft, the width is 4, and the height is 10. the volume is 40. the base would be 0.4 sq ft, costing 1.6 dollars. each of the sides would again be 10 sq ft, costing 20 dollars each. that gives me a total of $81.60 cant i keep making the length smaller and smaller for the price to keep going down? i do not understand! 4. Aug 22, 2007 ### omg precal Please, someone help me solve c, d, and e!!! i urgently and really really RRREEEEAAALLLYYY need this now! im begging please help me! 5. Aug 22, 2007 ### omg precal This Is Urgent!!! 6. Aug 22, 2007 ### learningphysics For part c, substitute l = 10/h into your cost equation in a. For part d, what values can h take for a valid cost? That's your domain. For part e, graph the equation in part c... find the h where the cost is minimum... then find l. w is already given as 4ft. 7. Aug 22, 2007 ### omg precal OMG you own life! thanks a lot man 8. Aug 22, 2007 ### learningphysics Are you sure you have the question written right? Like you said, it seems like the cost keeps dropping for increasing h, so I don't see a minimum cost. 9. Aug 22, 2007 ### omg precal thats exactly what i thought. my parents even thought that, and i came here in hope of a better response =p this is copied exactly from a precalculus summer assignment. I have put 0.00...0001 as my answer so far, idk what to do. thx so much again. 10. Aug 22, 2007 ### learningphysics Yeah, I think the people that wrote the question made a mistake... hope someone else on the forum here can verify... The function seems to be: cost = 80 + 160/h... which just keeps dropping for increasing h... So a bigger h gives a lower cost. 11. Aug 22, 2007 ### omg precal yeah... i spent a good hour trying to find some kind of non existant hint in the wording or just simply another solution, but i couldnt. i just hope my teacher made a mistake, i dont wanna do bad on it. 12. Aug 22, 2007 ### omg precal and wouldnt that also screw up my domain for H? or would it be all real numbers lol 13. Aug 22, 2007 ### learningphysics Actually I think the minimum cost is$80. That is when h = infinity l = 0 and w = 4. Those aren't realistic dimensions, but it's something. :P

14. Aug 22, 2007

### learningphysics

All real numbers > 0.

15. Aug 22, 2007

### omg precal

wouldnt 0 length mean it is 2-dimensional? infinity height... that would be tall :rofl:

I think I'll just leave it blank.

16. Aug 22, 2007

### chroot

Staff Emeritus
Whoa! You're making a mistake very early in this problem.

The box has four sides. Two of them have area h * l, and two of them have area 4 * h. The cost of the sides is $2 per square foot, so the cost of the sides is (2*h*l + 2*4*h)*$2 = 16h + 4hl.

The bottom of he box has area 4l, and costs \$4 per square foot. The cost of the bottom is thus 4*4*l, or 16l.

The sum of the sides and bottom is 16h + 4hl + 16l. You have the cost function wrong, so everything that follows is wrong, too.

- Warren

17. Aug 22, 2007

### learningphysics

I was mistakenly thinking of all 4 sides being the same area! Sorry omg_precal!

Last edited: Aug 23, 2007
18. Aug 23, 2007

### omg precal

Wow. I was blinded by stupidity. Thank you!!!

No problem at all. Thanks a lot for helping, though.

19. Aug 23, 2007

### omg precal

This is, by far, the hardest summer assignment I've ever had...

But thanks to you guys, I managed. =]

20. Aug 23, 2007

Cool!