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 square metre for the base and Rs. 45 per square metre for sides, what is the cost of least expensive tank?

Solution

Let l, b, and h denote the length breadth and depth of the open rectangular tank.

Given h = 2m

V = 8m³

i.e. 2 l b= 8

$\Rightarrow l b = 4 or b = \frac{4}{l}$

Surface area, S, of the open rectangular tank of the depth 'h' = l b + 2( l + b ) x h

In this problem, b = $\frac{4}{l}$ , l b = 4 metre, h = 2 metre

$∴ S = 4 + 2 (l + \frac{4}{l}) x 2 \Rightarrow S = 4 + 4 (l + \frac{4}{l})$

For maxima or minima, differentiating with respect to l we get,

$\frac{ds}{d l} = 4 (1 - \frac{4}{l^{2}}) \frac{ds}{d l} = 0 \Rightarrow l = 2 m$

l = 2m for minimum 0r maximum

$Now, \frac{d^{2} s}{d l^{2}} = \frac{48}{l^{3}} > 0 for all l So, l = 2 m is a point of minima and minimum surface area is S = l b + 2 (l + b) x h$

= 4 + 2 x 8 = 4 + 16 = 20 square metres

Base Area = 4 square metres; Lateral surface area = 16 square metres

Cost = 4 x 70 + 16 x 45

= 280 + 720

= Rs. 1000.

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