Let the length, width and height of the open tank be x, x and y units respectively. Then, its volume is and the total suface area is .
It is given that the tank can hold a given quantity of water. This means that its volume is constant. Let it be V.
The cost of the material will be least if the total surface area is least. Let S denote the total surface area.
We have to minimize S object to the condition that the volume V is constant.
Differentiating w.r.t x, we get,
Differentiating again w.r.t x, we get,
The critical numbers of S are given by
x = 2y
> 0, for all x
Hence, S is minimum when x = 2y, i.e. the depth (height) of the tank is half of its width.
Value – In the above mentioned question, the values that are being shown are of care, concern towards the weaker section of society.
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