Let the length, breath and height of tank be l, b and h respectively.
Also,
Assume volume of tank as V
h = 2 m (given)
V = 8 cm3
⇒ lbh = 8
⇒ 2lb = 8 (given)
⇒ lb =4
⇒ b = \(\frac{4}{1}\)…(1)
Cost for building base = Rs 70/m2
Cost for building sides = Rs 45/m2
Cost for building the tank, C = Cost for base + cost for sides
⇒ C = lb × 70 + 2(l + b) h × 45
⇒ C = l × \(\frac{4}{I}\) × 70 + 2(l + \(\frac{4}{I}\)) × 2 × 45
⇒ C = 280 + 180(l + \(\frac{4}{I}\)) …(2)
Condition for maxima and minima,
⇒ \(\frac{dC}{dI}\) = 0
⇒ 180(1 - \(\frac{4}{I^2}\)) = 0
⇒ \(\frac{4}{I^2}\) = 1
⇒ l2 = 4
⇒ l = ±2 cm
Since,
l cannot be negative
So,
l = 2 cm
\(\frac{d^2c}{dI^2}\) = 180\((\frac{8}{I^3})\)
For l = 2\(\frac{d^2c}{dI^2}\) = 180
⇒ \(\frac{d^2c}{dI^2}\) > 0
Therefore,
Cost will be minimum for l = 2
From equation 2,
C = 280 + 180(l +\(\frac{4}{I}\))
⇒ C = Rs 1000