How to Minimize Cost for an Open-Top Box with Given Volume and Width

[omg precal]

Homework Statement

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.

Homework Equations

NONE

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.

 

[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).

 

[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!

 

[omg precal]

Please, someone help me solve c, d, and e! i urgently and really really RRREEEEAAALLLYYY need this now! I am begging please help me!

 

[omg precal]

This Is Urgent!

 

[learningphysics]

Please, someone help me solve c, d, and e! i urgently and really really RRREEEEAAALLLYYY need this now! I am begging please help me!

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.

 

[omg precal]

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.

OMG you own life! thanks a lot man

 

[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.

 

[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.

 

[learningphysics]

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.

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.

 

[omg precal]

yeah... i spent a good hour trying to find some kind of non existent hint in the wording or just simply another solution, but i couldnt. i just hope my teacher made a mistake, i don't want to do bad on it.

 

[omg precal]

and wouldn't that also screw up my domain for H? or would it be all real numbers lol

 

[learningphysics]

yeah... i spent a good hour trying to find some kind of non existent hint in the wording or just simply another solution, but i couldnt. i just hope my teacher made a mistake, i don't want to do bad on it.

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

 

[learningphysics]

and wouldn't that also screw up my domain for H? or would it be all real numbers lol

All real numbers > 0.

 

[omg precal]

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

I think I'll just leave it blank.

 

[chroot]

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

 

[learningphysics]

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

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

 

[omg precal]

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

Wow. I was blinded by stupidity. Thank you!

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

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

 

[omg precal]

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

But thanks to you guys, I managed. =]

 

[learningphysics]

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

But thanks to you guys, I managed. =]

Cool!

 

[omg precal]

Cost Function = 16l + 16h + 4lh

Cost Function (h) = 160/h + 16h + 40

Height = 10
Length = 1
Width = 4

16(10) + 16(1) + 4(1)(10) = 216 dollars

16(100) + 16(0.1) + 4(0.1)(100) = 1641.60 dollars (smaller you go, much bigger the price gets!)

16(5) + 16(2) + 4(5)(2) = 156 dollars!

Now, let's say that the length and height are both sqrt(10)...

16(sqrt10) + 16(sqrt10) + 4(sqrt10)(sqrt10) = 141.20 dollars, which is the minimum, I believe.

If I'm wrong, correct me, please!

YAY!

 

[learningphysics]

Cost Function = 16l + 16h + 4lh

Cost Function (h) = 160/h + 16h + 40

Height = 10
Length = 1
Width = 4

16(10) + 16(1) + 4(1)(10) = 216 dollars

16(100) + 16(0.1) + 4(0.1)(100) = 1641.60 dollars (smaller you go, much bigger the price gets!)

16(5) + 16(2) + 4(5)(2) = 156 dollars!

Now, let's say that the length and height are both sqrt(10)...

16(sqrt10) + 16(sqrt10) + 4(sqrt10)(sqrt10) = 141.20 dollars, which is the minimum, I believe.

If I'm wrong, correct me, please!

YAY!

You got the right answer. :)

 

[omg precal]

Thanks!