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Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.

 

 

 

 
 
 
 
 

Answers (1)

Then volume of the cubiod \left ( v \right )= x^{2}y
                     \Rightarrow y= \frac{v}{x^{2}}---(i)
Now, surface area of cubiod
s= 2\left ( x^{2}+yx+yx \right )
\Rightarrow s= 2\left ( x^{2}+2xy \right )\Rightarrow s= 2\left ( x^{2}+2x\left ( \frac{v}{x^{2}} \right ) \right )\; \left [ using (i) \right ]
\Rightarrow s= 2\left ( x^{2}+\frac{2v}{x} \right )
Now differentaiting s w.r.t x we get
\frac{ds}{dx}= 2\left ( 2x-\frac{2v}{x^{2}} \right )
for maxima & minima , we have \frac{ds}{dx}= 0
\Rightarrow 2\left ( 2x-\frac{2v}{x^{2}} \right )= 0\, \Rightarrow 4\left ( x-\frac{v}{x^{2}} \right )= 0
x= \frac{v}{x^{2}}\Rightarrow x= \sqrt[3]{v}
Now \frac{ds}{dx}= 2\left ( 2x-\frac{2v}{x^{2}} \right )= 4\left ( x-\frac{v}{x^{2}} \right )
Again differentiating w.r.t x we get
\frac{d^{2}s}{dx^{2}}= 4\left [ 1+\frac{2v}{x^{3}} \right ]
\Rightarrow \left [ \frac{d^{2}s}{dx^{2}} \right ]_{x= 3\sqrt{v}}= 4\left [ 1+\frac{2v}{v} \right ]= 12> 0
Thus, s is minimum when x= 3\sqrt{v}
putting x= \sqrt[3]{v}  in (i) we get
y= \frac{v}{x^{2}}= \frac{x^{3}}{x^{2}}= x\Rightarrow y= x
Hence it is a cube since the length, breadth and height of a cube are equal. Hence proved.

Posted by

Ravindra Pindel

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