An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when the depth of the tank is half of its width. If the cost is to be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question?
Let the length, width and height of the open tank be x, x and y units respectively. Then, its volume is x2Y and the total surface area is x2 + 4xy.
It is given that the tank can hold a given quantity of water. This means that its volume is constant. Let it be V. Then,
V = x2y
The cost of the material will be least if the total surface area is least. Let S denote the total surface area.
Then,
S = x2 + 4xy
We have to minimize S object to the condition that the volume V is constant.
Now,
The critical numbers of S are given by dS/dx = 0
Now, ds/dx = 0
Hence, S is minimum when x =2y i.e the depth (height) of the tank is half of its width.
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