Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

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?

1 Answer

0 votes
  • 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
Let the length of the rectangle be = $x$
Let the breadth of the rectangle be = $y$
$ \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$
$ \therefore \bigg( \large\frac{d^2v}{dx^2}\bigg)_{(12, 6)} = 36-6 \times 12$
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 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App