# Math Help - area & volume of an open box

1. ## 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 2. 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?

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

3. Hello, confusedagain!

The Captain made a teensy error . . .

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