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Solution for A Tank with Rectangular Base and Rectangular Sides, Open at the Top is to the Constructed So that Its Depth is 2 M and Volume is 8 M3. If Building of Tank Cost 70 - CBSE (Science) Class 12 - Mathematics

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Question

A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is 2 m and volume is 8 m3. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank?

Solution

Let lb and h be the length, breadth and height of the tank, respectively.

Height, h = 2 m

Volume of the tank = 8 m3

Volume of the tank = l × b × h

∴  × b × 2 = 8

\[\Rightarrow lb = 4\]

\[ \Rightarrow b = \frac{4}{l}\]

Area of the base = lb = 4 m2

Area of the 4 walls, A= 2h (l + b)

\[\therefore A = 4\left( l + \frac{4}{l} \right)\]

\[ \Rightarrow \frac{dA}{dl} = 4\left( 1 - \frac{4}{l^2} \right)\]

\[\text { For maximum or minimum values of A, we must have }\]

\[\frac{dA}{dl} = 0\]

\[ \Rightarrow 4\left( 1 - \frac{4}{l^2} \right) = 0\]

\[ \Rightarrow l = \pm 2\]

However, the length cannot be negative.

Thus,
l = 2 m

\[\therefore b = \frac{4}{2} = 2 m\]

\[\text { Now,} \]

\[\frac{d^2 A}{d l^2} = \frac{32}{l^3}\]

\[\text { At }l = 2: \]

\[\frac{d^2 A}{d l^2} = \frac{32}{8} = 4 > 0\]

Thus, the area is the minimum when l = 2 m

We have
l = b = h = 2 m

∴ Cost of building the base = Rs 70 × (lb) = Rs 70 × 4 = Rs 280

Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2)= Rs 8 (90) = Rs 720

Total cost = Rs (280 + 720) = Rs 1000

Hence, the total cost of the tank will be Rs 1000.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.

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Solution A Tank with Rectangular Base and Rectangular Sides, Open at the Top is to the Constructed So that Its Depth is 2 M and Volume is 8 M3. If Building of Tank Cost 70 Concept: Graph of Maxima and Minima.
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