Question

A tank with a rectangular base and rectangular sides, open at the top is made. Depth of the tank is \(4\) m and its volume is \(36\) cubic meters. For making a tank cost of base material used is Rs. \(100\) per sq. meter and that of sides is Rs. \(50\) per sq. meter. Then minimum cost of tank is ..............

• A. Rs. \(1100\)
• B. Rs. \(2200\)
• C. Rs. \(3300\)
• D. Rs. \(4400\)

Hint:

Fix the base area from the volume, then minimise the cost of the sides using AM-GM.

Answer 1

The Correct Option is C

Step 1: Calculus Version:
Let the base be $l$ by $9/l$. The cost is $C(l)=900+400\left(l+\dfrac9l\right)$.

Step 2: Differentiate:
$C'(l)=400\left(1-\dfrac{9}{l^2}\right)=0\Rightarrow l=3$. $C''(l)=\dfrac{7200}{l^3}>0$, so it is a minimum.

Step 3: Value:
At $l=3$: $l+9/l=6$, so $C=900+2400=3300$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) Rs. } 3300} \]