# Thread: Box Volume/Calculus problem

1. ## Box Volume/Calculus problem

Trying to find the maximum volume achievable for a container to be made from a steel sheet measuring 3.4m X 1.9m.

This is my workings so far,where am i going wrong?

v = l x w x h

V(h) (304 - 2h)(1.9 - 2h) X h
(6.46 - 13.6h + 4h^2) x h
4h^3 - 13.6h^2 + 6.46h
v(h) 4h^3 - 13.6h^2 + 6.46h
v(h) 12h^2 - 27.2h + 6.46
0 = 12h^2 - 27.2h + 6.46 divide both sides by 12
0 = 2.266 + 0.538 = 2.804 I Think I am supposed to factor around here!
h = 2.804
V = 4h^3 -27.2 + 6.46 = 1390.2

Any advice appreciated

Thanks.....

2. Originally Posted by manich44
Trying to find the maximum volume for a container to be made from a steel sheet measuring 3.4m X 1.9m.

This is my workings so far,where am i going wrong?

v = l x w x h

V(h) (304 - 2h)(1.9 - 2h) X h
(6.46 - 13.6h + 4h^2) x h
4h^3 - 13.6h^2 + 6.46h
v(h) 4h^3 - 13.6h^2 + 6.46h
v(h) 12h^2 - 27.2h + 6.46
0 = 12h^2 - 27.2h + 6.46 divide both sides by 12
0 = 2.266 + 0.538 = 2.804 I Think I am supposed to factor around here!
h = 2.804
V = 4h^3 -27.2 + 6.46 = 1390.2

Any advice appreciated

Thanks.....
Your work is quite "messy".
Try to be clearer: keep your teacher in good mood

Multiply dimensions by 10: 34 by 19; easier to handle.

V = h(19 - 2h)(34 - 2h)
V = 4h^3 - 106h^2 + 646h ; divide by 2:
V = 2h^3 - 53h^2 + 323h
Still with me? OK:
6h^2 - 106h + 323 = 0
Solve using quadratic; you'll get h = ~13.75 or ~3.92

13.75 impossible, so h = 3.92; or .392 due to original multiplication by 10

So h = .392, w = 1.9 - 2(.392), l = 3.4 - 2(.392)

Finish it...

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