Differentiation 03

The open box consists of 4 sides and one base
Let b be the length of the base and h be the height of the box
Area of base = b^2
Total area of 4 sides = 4bh
Total cost of the box = cost of base + cost of 4 sides = $4(b^2) + $3(4bh)
$48 = $4b^2 + $12bh
48 - 4b^2 = 12bh
3bh = 12 –b^2
h = (12 –b^2)/(3b)
Volume = hb^2
V =b^2 (12 –b^2)/(3b)
V =b (12 –b^2)/3
V = 4b – (1/3)b^3
Take first derivative: Differentiate V with respect to b
dV/db = 4 –b^2
0 = 4 –b^2
b^2 = 4
b = 2
h = (12 –b^2)/(3b)
h = (12 –4)/6
h =8/6 4/3
Take second derivative
d2v/db2 = -2b
d2v/db2 = -2(2) = -4 Negative indicates maximum

The box is 4/3 m high and the side length of the square base is 2 m (answer)

Check the cost:
Cost of Base = $4/sq. m x 4 sq. m = $16
Cost of 4 sides = 4 x $3/sq. m x (4/3 m) (2 m) = $32
Total cost = $16 + $32 = $48

let b, h be the base and height of box
Volume = hb^2
cost for sides = 4bh/3
cost for base = b^2/4
4bh/3 + b^2/4 = 48
16bh + 3b^2 = 576
16hb^2 + 3b^3 = 576b
16V + 3b^3 = 576b
V = (1/16)(576b - 3b^3)
dV/db = (1/16)(576 - 9b^2) = 0
second derivatives = -18b which is < 0 and maximise V when dV/db = 0
576 - 9b^2 = 0 => b = 8 and h = 3