optimization problem

**raymac62** · October 2nd 2011, 07:08 PM

I am totally lost with this problem.

The length of a cedar chest is twice its width. The cost/dcm^2 of the lid is four times the cost/dcm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440 dcm^3, find the dimensions so that the cost is a minimum.

**mr fantastic** · October 2nd 2011, 08:18 PM

Originally Posted by raymac62

I am totally lost with this problem.

The length of a cedar chest is twice its width. The cost/dcm^2 of the lid is four times the cost/dcm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440 dm^3, find the dimensions so that the cost is a minimum.

C = 4(2x^2) + 2x^2 + 2(2xh) + 2(xh) = .....

where 1440 = 2x^2h.

Therefore C(x) = .......

and proceed in the usual way.

If you need more help, please show all your work and say where you get stuck.

**CaptainBlack** · October 3rd 2011, 05:55 AM

Originally Posted by raymac62

I am totally lost with this problem.

The length of a cedar chest is twice its width. The cost/dcm^2 of the lid is four times the cost/dcm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440 dm^3, find the dimensions so that the cost is a minimum.

What are dcm^2 and dm^3?

CB

**raymac62** · October 3rd 2011, 07:28 AM

They are typos. I was very tired when I made the original post. I should have wrote "dcm" in both cases. Even on the diagram I have labelled the cedar chest to have a a volume in cubic centimeters where it should be decimeters. 1440dcm cubed
One cubic decimeter is: 10cm * 10cm * 10cm

** Of course I have edited it now.

**raymac62** · October 3rd 2011, 08:02 AM

Originally Posted by mr fantastic

C = 4(2x^2) + 2x^2 + 2(2xh) + 2(xh) = .....

where 1440 = 2x^2h.

Therefore C(x) = .......

and proceed in the usual way.

If you need more help, please show all your work and say where you get stuck.

C = 11x^2 + 6hx + h^2

Why is the volume expressed: 2x^2h? Should it not be: 2x^2 * h ?

**mr fantastic** · October 3rd 2011, 12:55 PM

Originally Posted by raymac62

C = 11x^2 + 6hx + h^2

Why is the volume expressed: 2x^2h? Should it not be: 2x^2 * h ?

The implied multiplication ought to be crystal clear to you from the context. What I posted was help, not a pristine solution.

**raymac62** · October 3rd 2011, 03:58 PM

Originally Posted by mr fantastic

The implied multiplication ought to be crystal clear to you from the context. What I posted was help, not a pristine solution.

Let me explain what I know (or at least think I know) so that there is a better understanding of context:

C = the total cost for producing the cedar chest.

V = 1440 - In someway I have to take this value and substitute it into the equation so I can determine the value of at least 1 variable.

Then, I have to find the derivative, and solve for x when it is equal to zero.

I am not sure how I can go about expressing the total cost with only a single variable.

---
Obviously, what you posted was not a pristine solution. Nonetheless, why have you expressed the volume as 2x^2h? That makes absolutely no sense to me. I do not see how that is algebraically equivalent to expressing V as 2x^2 * h. However, it must mean something to you. If you would care to explain...

Thanks in advance.

Sincerely,

Raymond

**mr fantastic** · October 3rd 2011, 08:40 PM

Originally Posted by raymac62

Let me explain what I know (or at least think I know) so that there is a better understanding of context:

C = the total cost for producing the cedar chest.

V = 1440 - In someway I have to take this value and substitute it into the equation so I can determine the value of at least 1 variable.

Then, I have to find the derivative, and solve for x when it is equal to zero.

I am not sure how I can go about expressing the total cost with only a single variable.

---
Obviously, what you posted was not a pristine solution. Nonetheless, why have you expressed the volume as 2x^2h? That makes absolutely no sense to me. I do not see how that is algebraically equivalent to expressing V as 2x^2 * h. However, it must mean something to you. If you would care to explain...

Thanks in advance.

Sincerely,

Raymond

V = (2)*(x^2)*(h).

Better now? (I guess in the age of the CAS calculator, implied multiplication is not as clearly understood is it once was).

1440 = 2x^2h => h = 720/x^2. Substitute into C etc. etc.

**raymac62** · October 5th 2011, 12:47 PM

Originally Posted by mr fantastic

V = (2)*(x^2)*(h).

Better now? (I guess in the age of the CAS calculator, implied multiplication is not as clearly understood is it once was).

1440 = 2x^2h => h = 720/x^2. Substitute into C etc. etc.

This is what I have and somethings not working out.
Could you please show me where I went wrong.
Thanks in advance!

$C=4(2x^2) + 2x^2 + 2(2hx) + 2(hx)=11x^2+6hx+h^2$

$h= \frac {1440}{2x^2}$

$C(x)=11x^2 + \frac {6(1440x)}{x^2} + [\frac {1440}{x^2}]^2$ ...sorry don't know large bracket command

$= 11x^2 + \frac {8640}{x} + \frac {2073600}{x^4} = 11x^2 + 8640(x)^{-1} + 2073600(x)^{-4}$

$C'(x)=22x - \frac {8640}{x^2} - \frac {8294400}{x^5}$

Solve for x when the derivative is equal to zero.

$22x^6 = \frac {8640x^5}{x^2} + 8,294,400$

$22x^6 -8640x^3 - 8294400 = 0$

$x \approx 9.44$

Dimensions obtained: W= 9.44 L=18.88 h=16.16

These don't work out to equal 1440, So I know I am going wrong somewhere. Can't seem to pinpoint exactly where.

Once again, thanks a lot for your help.

Sincerely,

Raymond

**raymac62** · October 5th 2011, 12:52 PM

Think i have spotted where I went wrong. 2(hx) should be (h^2)(x^2) where I went h^2 +2hx +x^2 in the original Cost equation.

**mr fantastic** · October 5th 2011, 01:12 PM

Originally Posted by raymac62

Think i have spotted where I went wrong. 2(hx) should be (h^2)(x^2) where I went h^2 +2hx +x^2 in the original Cost equation.

The above makes no sense to me. Neither does your expansion and simplification in your previous post.

I would have thought that after expanding, $C = 8x^2 + 2x^2 + 4hx + 2hx = ....$

so I'm not sure how you got your result ....

**raymac62** · October 6th 2011, 03:08 PM

Looks like I need to review my algebra. :P I'm over complicating things. I got it now.

2(xy) is not equal to (x^2)(y^2).... I was drained.

**raymac62** · October 6th 2011, 03:47 PM

Originally Posted by mr fantastic

The above makes no sense to me. Neither does your expansion and simplification in your previous post.

I would have thought that after expanding, $C = 8x^2 + 2x^2 + 4hx + 2hx = ....$

so I'm not sure how you got your result ....

$C=4(2x^2) + 2x^2 + 2(2hx) + 2(hx) = 10x^2 + 6hx$

$h= \frac {1440}{2x^2} = \frac {720}{x}$

$C(x)=10x^2 + \frac {6(720)x}{x^2}$ .

$= 10x^2 + \frac {4320}{x} = 10x^2 + 4320(x)^{-1}$

$C'(x)=20x - \frac {4320}{x^2}$

Solve for x when the derivative is equal to zero.

$20x^3 = 4320$

$x^3 = 216$

$x = 216^{1/3}$

$x = 6$

Thus, $W=6, L=12, H=20$

6*12*20= 1440 ...yay.

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