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ISI2018-DCG-21

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A box with a square base of length $x$ and height $y$ has an open top and its volume is $32$ cubic centimetres, as shown in the figure below. The values of $x$ and $y$ that minimize the surface area of the box are

  1. $x=4$ cm $\&$ $y=2 $ cm
  2. $x=3$ cm $\&$ $y=\frac{32}{9} $ cm
  3. $x=2$ cm $\&$ $y=8 $ cm
  4. none of these.

2 Answers

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if the base length is x and height is y

the volume= x^2y=32 so, y=32/x^2

now, the surface area= (area of x sided square){ below}+4xy(rectangular side with length y, breadth x)

=x^2+4xy= x^2+4x*32/x^2= x^2+128/x

so, after first derivative

f'(x)=2x-128/x^2

f'(x)=0 at x=4

f''(x)=2+256/x^3 (second derivative)

the second derivative is positive at x=4

so, the function has a minima at x=4,

y=32/x^2=32/16=2

so, the surface area will be minimum at x=4 and y=2
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