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April 18, 2014

April 18, 2014

Posted by **Andrew** on Monday, July 18, 2011 at 1:42pm.

- Calculus -
**Reiny**, Monday, July 18, 2011 at 2:17pmLet each of the cut-outs be x

then the base of the box is 24 - 2x

the width of the box is 12-2x

and its height is x

Volume = V = x(24-2x)(12-2x)

expand, differentiate and set equal to zero

you will be solving a quadratic.

- Calculus -
**Andrew**, Monday, July 18, 2011 at 2:48pmWould that make the dimensions x=6 and 12? or did I do that wrong?

- Calculus -
**Reiny**, Monday, July 18, 2011 at 3:04pmHow did you possible get these answers ?

They make no sense.

If x = 6, then you would be cutting away 12 cm from the base of width 12, so nothing is left.

x = 12 would be physically impossible, you would be cutting away 24 cm from the 12 cm width !

Did you even differentiate ?

Show me what you did.

- Calculus -
**Andrew**, Monday, July 18, 2011 at 3:09pmi got V= 4x^3-72x^2+288x

V=12x^2-144x+288=0

Then I foiled that and my answers were x=6 and x=12. Clearly I have no idea what I'm doing, I have a test in my class on wednesday and I need to be able to make sense of this in my homework due tomorrow

- Calculus -
**Reiny**, Monday, July 18, 2011 at 3:26pmYour equation 12x^2-144x+288=0 is correct

from there, divide each term by 12

x^2 - 12 + 24 = 0

This does not factor (FOIL is used to expand, not in factoring)

so ...

x = (12 ± √(144 - 4(1)(24))/2

= (12 ± √48)/2

= (12 ± 4√3)/2 = 6 ± 2√3

= aprr. 2.536 or 9.464

but clearly x < 6 or else the width would become negative, so

x = 2.536

length = 24-2(2.536) = 19.928

width = 12 - 2(2.536) = 6.928

height = 2.536

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