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  • 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 depth of the tank is half of its width. If the cost is to be borne by

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 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 ?

 

 

 

 
 
 
 
 

Answers (1)

Let the length breadth and height of the open tank be x, and y units respectively.
Then Volume \left ( v \right )= x^{2}y---(i)
Total surface area \left ( s \right )= x^{2}+4xy---(ii)
                        s= x^{2}+4x\frac{v}{x^{2}}---(using\, i)
\Rightarrow \frac{ds}{dx}= 2x-\frac{4v}{x^{2}}
For critical points, put
\frac{ds}{dx}= 0
\Rightarrow 2x-\frac{4v}{x^{2}}= 0
\Rightarrow 2x^{3}= 4v
\Rightarrow 2x^{3}= 4x^{2}y\; \; \left [ using\left ( i \right ) \right ]
\Rightarrow x= 2y---(iii)
Now, \frac{d^{2}s}{dx^{2}}= 2+\frac{8v}{x^{3}}
                 = 2+\frac{8v}{8y^{3}}\; \; \left [ using (iii) \right ]
               \Rightarrow 2+\frac{v}{y^{3}}> 0
The area is minimum, thus cost is minimum when x = 2y
ie. depth of the tank is half of the width.

Posted by

Ravindra Pindel

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