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# Find the dimentions of the rectangle of perimeter 36cm which will sweep out a volume as large as possible,when revolved about one of its sides.Also find the maximum volume.

Can you answer this question?

Toolbox:
• 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 $f''(x)>0$
Step 1
Perimeter of the rectangle is
$p=36$
Let the length of the rectangle be = $x$
Let the breadth of the rectangle be = $y$
$p=2(x+y)=36$
$\Rightarrow x+y=18 \Rightarrow y=18-x$
As the rectangle revolves, the
length = $x$
width = $18-x$
height = $x$
$\therefore$ volume of the cubid is
$v = x \times (18-x) \times x$
$= x^2(18-x)$
$= 18x^2-x^3$
Step 2
Differentiating w.r.t $x$ we get
$\large\frac{dv}{dx}=36x-3x^2$
differentiating w.r.t $x$ we get,
$\large\frac{d^2v}{dx^2}=36-6x$
Step 3
For maximum or minimum we must have
$\large\frac{dv}{dx}=0$
$\Rightarrow 36x-3x^2=0$
$\Rightarrow 3x=36$
$x=12$
$\therefore y=18-12$
$=6$
$\therefore \bigg( \large\frac{d^2v}{dx^2}\bigg)_{(12, 6)} = 36-6 \times 12$
$=36-72$
$-36<0$
Hence $v$ is maximum when $x=12$
Step 4
$\therefore$ The dimensions of the rectangle is
$x=12\: and \: y=6$
The maximum volume of the box is given by
$v=12 \times 12 \times 6$
$= 864$ cubic units
answered Aug 8, 2013
Incorrect answer. When a rectangle revolves around a side then it forms a cylinder