A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?
The cost of the least expensive tank is Rs 1000.
Explanation
This is an optimization problem from the Application of Derivatives chapter. We need to find the dimensions that minimize the total cost of constructing the tank. The key is to express the cost as a function of one variable using the volume constraint, then find the minimum using derivatives.
Solution Steps
Step 1: Let the length and breadth of the rectangular base be x m and y m respectively. Given depth = 2 m and Volume = 8 m³.
Step 2: Volume = x × y × 2 = 8, therefore xy = 4 or y = 4/x.
Step 3: Calculate areas - Base area = xy = 4 m². Area of four sides = 2(2x) + 2(2y) = 4x + 4y = 4(x + y) m².
Step 4: Express total cost C = Cost of base + Cost of sides = 70(4) + 45(4)(x + y) = 280 + 180(x + y).
Step 5: Substitute y = 4/x: C = 280 + 180(x + 4/x). To minimize C, we minimize f(x) = x + 4/x.
Step 6: Find f'(x) = 1 - 4/x². Setting f'(x) = 0 gives x² = 4, so x = 2 (positive value).
Step 7: Verify minima: f''(x) = 8/x³. At x = 2, f''(2) = 1 > 0, confirming minima.
Step 8: When x = 2, y = 4/2 = 2 m. Minimum cost = 280 + 180(2 + 2) = 280 + 720 = Rs 1000.