Two variables need to equal a product

• March 9th 2008, 05:09 AM
No Logic Sense
Two variables need to equal a product
I need to construct a train cart.

The train must have a volume of $1.25 m^3$

As a train cart is on tracks, the width is set to 0.8 meters and cannot change obviously.

That leaves the length and height of the cart to be the only variables we can change:

$(0.8 * a * b)m = 1.25 m^3$

A quick reduction where 0.8 is divided on both sides gives:

$(a * b)m = 1.5625 m^2$

So, now my question is, how do I decide what the two variables (the length and height of the cart) should be?

And there is one more last thing to the question. Since the material for the cart is very expensive, you must use as little material (the surface areal of the cart) as possible. The cart is obviously open at the top of the cart.
• March 9th 2008, 06:24 AM
Moo
Hi,

Calculate the total area of the cart with a,b and 0.8. Then find the minimum...

Am not sure it'll give something, but try
• March 9th 2008, 06:33 AM
earboth
Quote:

Originally Posted by No Logic Sense
I need to construct a train cart.

The train must have a volume of $1.25 m^3$

As a train cart is on tracks, the width is set to 0.8 meters and cannot change obviously.

That leaves the length and height of the cart to be the only variables we can change:

$(0.8 * a * b)m = 1.25 m^3$

A quick reduction where 0.8 is divided on both sides gives:

$(a * b)m = 1.5625 m^2$

So, now my question is, how do I decide what the two variables (the length and height of the cart) should be?

And there is one more last thing to the question. Since the material for the cart is very expensive, you must use as little material (the surface areal of the cart) as possible. The cart is obviously open at the top of the cart.

$l \cdot h = \frac{25}{16}~\implies~l = \frac{25}{16h}$

Calculate the surface area:

$s = l\cdot h + 2 \cdot 0.8 \cdot h + 2 \cdot 0.8 \cdot l$

$s = \frac{25}{16} + 1.6 \cdot h + 1.6 \cdot l$ . that means:

$s(h)=\frac{25}{16} + 1.6 \cdot h + 1.6 \cdot \frac{25}{16h}$

You get the extreme(minimum or maximum) value of s(h) if s'(h) = 0. So:

a) Calculate the first derivative of s.
b) Solve the equation s'(h) = 0 for h.
c) Plug in the value of h into the term calculating l.

I've got $l = h = \frac54\ m$
• March 9th 2008, 09:29 AM
No Logic Sense
Thanks for the replies!

There's one thing I think you've done wrong however. I think you used a wrong equation for the surface. I believe it should be like this:

$s = 0.8 * l + 2 * 0.8 * h + 2 * h * l$

As the top of the cart is open, and the top of the carts 2 dimensions are 0.8 * the length, it should be this dimension that doesn't get timed with 2, right?

So since $h * l = 25/16$ and $l = 25/16h$, we will get:

$s = 0.8 * 25/16h + 1.6 * h + 2 * 25/16$
$s(h) = 0.8 * 25/16h + 1.6 * h + 50/16$

Right?

So the derivative of s'(h) will then be:

$s'(h) = (25/16h)^{-1} + 1.6$

I think this is wrong though? :(
• March 9th 2008, 10:34 PM
earboth
Quote:

Originally Posted by No Logic Sense
Thanks for the replies!

There's one thing I think you've done wrong however. I think you used a wrong equation for the surface. I believe it should be like this:

$s = 0.8 * l + 2 * 0.8 * h + 2 * h * l$

As the top of the cart is open, and the top of the carts 2 dimensions are 0.8 * the length, it should be this dimension that doesn't get timed with 2, right? ... Yes

So since $h * l = 25/16$ and $l = 25/16h$, we will get:

$s = 0.8 * 25/16h + 1.6 * h + 2 * 25/16$
$s(h) = 0.8 * 25/16h + 1.6 * h + 50/16$

Right?... Yes

So the derivative of s'(h) will then be:

$s'(h) = (25/16h)^{-1} + 1.6$... no

I think this is wrong though? :(

I rewrite the equation of s:

$s(h) = 0.8 \cdot \frac{25}{16 \cdot h} + 1.6 h + \frac{50}{16} = \frac{5}{4} \cdot h^{-1} + \frac85 h + \frac{25}{8}$ . The first derivative of s is:

$s'(h) = -\frac54 \cdot h^{-2} + \frac85$ . So if s'(h) = 0 you'll get:

$-\frac54 \cdot h^{-2} + \frac85 = 0~\iff~ \frac54 \cdot h^{-2} = \frac85 ~\iff~ h^2 = \frac{25}{32}~\implies~ h = \frac{5}{8} \cdot \sqrt{2}$
• March 10th 2008, 08:48 AM
No Logic Sense
Alright, thanks for the help. :) I get it now. :)