# area & volume of an open box

• Mar 12th 2007, 07:22 PM
confusedagain
area & volume of an open box
A carpenter has been asked to build an open box with a square base. The sides of the box will cost \$3 per square meter and the base will cost \$4 per square meter. What are the dimensins of the box of greatest volume that can be constructed for \$48?

Should I start with finding the Surface Area or the Volume?
I'm not sure if I remember the formulas:
SA = 2b^2 + 4bh
V = b^2h
• Mar 12th 2007, 09:57 PM
CaptainBlack
Quote:

Originally Posted by confusedagain
A carpenter has been asked to build an open box with a square base. The sides of the box will cost \$3 per square meter and the base will cost \$4 per square meter. What are the dimensins of the box of greatest volume that can be constructed for \$48?

Should I start with finding the Surface Area or the Volume?
I'm not sure if I remember the formulas:
SA = 2b^2 + 4bh
V = b^2h

Let the side of the box be b, and the height be h, then the volume is:

V = b^2 h,

and the cost is:

C = 4 (b^2) + 3 (4 b h) = 4 b^2 +12 b h = 48

divide through by 4:

b^2 + 3 b h = 12.

So:

h = (12 - b^2)/(3 b)

Hence:

V = b^2 (12 - b^2)/(3 b) = b (12 - b^2)/3 = 4 b - b^3/3

The maximum volule will occur when dV/db=0, or:

dV/db = 4 - b^2 = 0

which has roots b=+/-2. The second derivative test tells us that b=+2
gives a maxima (that is d^2V/db^2<0 at b=2).

So b=2, then as h = (12 - b^2)/(3 b), h = 4/3

RonL
• Mar 13th 2007, 08:48 AM
Soroban
Hello, confusedagain!

The Captain made a teensy error . . .

Quote:

Originally Posted by CaptainBlack

V .= .b²(12 - b²)/(3b) .= .b(12 - b²)/3 .= .4b - b³/3

The maximum volume will occur when dV/db = 0, or:
. . dV/db .= .4 - b² .= .0

Therefore: .b = 2, h = 4/3