Q87 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
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 ..............
- ARs. $1100$
- BRs. $2200$
- Correct answerCRs. $3300$
- DRs. $4400$
Solution
Let the length and breadth of the base of the tank be $x$ and $y$ respectively.
Given depth $h = 4$ m and volume $V = 36$ m$^3$.
$V = x \times y \times h \Rightarrow 4xy = 36 \Rightarrow xy = 9$
Cost of the base $= 100 \times xy = 100 \times 9 = 900$
Area of the four sides $= 2(xh + yh) = 2(4)(x + y) = 8(x + y)$
Cost of the sides $= 50 \times 8(x + y) = 400(x + y)$
Total cost $C = 900 + 400(x + y)$
To minimize $C$, we need to minimize $x + y$.
Using AM-GM inequality for $x, y > 0$:
$\dfrac{x + y}{2} \geq \sqrt{xy}$
$x + y \geq 2\sqrt{9} = 6$
Minimum value of $x + y$ is $6$, which occurs when $x = y = 3$.
Minimum cost $C = 900 + 400(6) = 900 + 2400 = 3300$
Answer: Rs. $3300$
Given depth $h = 4$ m and volume $V = 36$ m$^3$.
$V = x \times y \times h \Rightarrow 4xy = 36 \Rightarrow xy = 9$
Cost of the base $= 100 \times xy = 100 \times 9 = 900$
Area of the four sides $= 2(xh + yh) = 2(4)(x + y) = 8(x + y)$
Cost of the sides $= 50 \times 8(x + y) = 400(x + y)$
Total cost $C = 900 + 400(x + y)$
To minimize $C$, we need to minimize $x + y$.
Using AM-GM inequality for $x, y > 0$:
$\dfrac{x + y}{2} \geq \sqrt{xy}$
$x + y \geq 2\sqrt{9} = 6$
Minimum value of $x + y$ is $6$, which occurs when $x = y = 3$.
Minimum cost $C = 900 + 400(6) = 900 + 2400 = 3300$
Answer: Rs. $3300$
Hence, the correct answer is option: 'C'.