height

ninjaman · 2014-02-24 07:56:37

hello

I have to find height of a box, I have length and width but I don't know height. unfortunately im not gifted with these things. my brain stops ticking. anyone got any tips or hints
thanks
simon

Bob · 2014-02-24 08:05:46

hi Simon,

If you know the volume then

If you don't know the volume then I'm at a loss.

Bob

ninjaman · 2014-02-28 08:53:39

NEVERMIND, I GOT THE ANSWER THANKS ANYWAY ALL THE BEST!!!:D:D

hello

I have this question

a rectangular sheet of metal measuring 120mm by 80mm has equal squares of size x cut from each of the corners. the remaining flaps are then folded upwards to form an open tray. draw a neat sketch of the net and prove that the volume of the tray is given by:

v = 4x^3 - 400x^2 + 9600x

a) find the value of x such that the volume is a maximum, making sure you show how you prove your value is a maximum not a minimum. work out the maximum volume of the tray.

b) verify your result for question 6a using a second method. present your alternative method in writing, ensuring you compare both ease of use and accuracy of results of the two methods.

the second bit is whats getting me. I can use the results of length and width which are 88.612mm and 48.612mm respectively. I cant include height and have to find volume.

im guessing its something to do with the cut outs.
I have maximum volume of box when x = 15.594

length = 120 -2 * 15.694 = 88.612
width = 80 -2 * 15.694 = 48.612

so I have the starting width and length, the finishing width and length and two lots of something to be taken away.

I found something close, I have 15.69499126 = x, this came from the general quadratic equation

I did this,

120 - 2 * 15.69499126 = 88.61001748

88.61001748 + 2 = 90.61001748

120 - 90.61001748 = 29.38998252

divide this by 2 to get = 14.69499126

this last number is supposed to be x but it is a whole 1 out. 14 instead of 15. I think this is correct but there is a 1 missing somewhere.
I think if I put all this together I should be able to get the correct answer giving me a final volume and height.
unless I have done something really daft here and got a knot in my brain.....again!:D

NEVERMIND, I GOT THE ANSWER THANKS ANYWAY ALL THE BEST!!!

anyway, if anyone could point me in the right direction. I would really appreciate it

thanks
simon

Last edited by ninjaman (2014-02-28 09:43:42)

Bob · 2014-02-28 21:06:42

hi Simon,

How did you get x? I get that answer too using calculus.

ie. work out dV/dx and put that equal to zero. This leads to a quadratic with two possible values for x.

There are ways to show whether a turning point (where dV/dx = 0) is a maximum or a minimum and using this you find there's one of each. Also you can reject one answer as it has a value of x that would not be possible with this sheet of metal. I can give more details on this if you want.

To get the max volume just do length x width x height using your value of x.

The question asks for a second method. I would use a graph plotter and zoom in to the maximum point. Calculus gives the exact answer straight away whereas the graph method is only as accurate as the plotter allows. And it takes time to zoom in sufficiently. If software is not permitted then you can still do a graph by hand. This gives you more to say, because it would not be anywhere near as accurate as the calculus method. You could even use trial and improvement which would be very slow.

Bob

Math Is Fun Forum

#1 2014-02-24 07:56:37

height

#2 2014-02-24 08:05:46

Re: height

#3 2014-02-28 08:53:39

Re: height

#4 2014-02-28 21:06:42

Re: height

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