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An open box with square base is to be made of a given quantity of card board of area $ 36 sq.metre.$ Find the maximum volume of the box.

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  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $ f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
Since the box is of square base,
Let the length, breadth and height of the box be $x, x\: and \: y$ units respectively.
Given: Area of the box = $36$ sq. metres.
We know that area of the open box=$x^2+4xy$
$ \therefore x^2+4xy=36 \Rightarrow y=\large\frac{36-x^2}{4x}$
Let $V$ be the volume of the box, then
$ \Rightarrow V=x^2 \bigg(\large\frac{36-x^2}{4x}\bigg)$
Differentiate w.r.t $x$ we get,
$\Rightarrow\: \large\frac{dV}{dx}$$=9-\large\frac{3x^2}{4}$
Differentiate again $\large\frac{dV}{dx }$ w.r.t $x$ we get,
$\Rightarrow\: \large\frac{d^2V}{dx^2}=-\large\frac{3x}{2}$
For maximum or minimum , we must have,
$ \large\frac{dV}{dx}$$=0 \Rightarrow 9-\large\frac{3x^2}{4}$$=0$
$ \Rightarrow \large\frac{3x^2}{4}$$=9$
$ \Rightarrow x=2\sqrt 3$
$ \bigg( \large\frac{d^2V}{dx^2} \bigg)_{x=2\sqrt 3}$$=-3\sqrt 3 <0$
Thus, $V$ is maximum when $x=2\sqrt 3$
Put $x=2\sqrt 3$, we get
$ y=\sqrt 3$
$ \therefore$ The maximum volume of the box is given by
$ V= x^2y=12\sqrt 3$ cubic metres.
answered Feb 4, 2014 by rvidyagovindarajan_1
edited Feb 4, 2014 by rvidyagovindarajan_1

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