logo

Ask Questions, Get Answers

X
 
Home  >>  CBSE XII  >>  Math  >>  Model Papers

A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.

1 Answer

 
Second Derivative Test - If f'(x) = 0 at a point and f''(x) > 0 at this point, then f(x) has a local minimum at this point.
 
Let $s$ be the side of square and $a$ be side of equilateral triangle.
 
Given,
 
$4s + 3a = 36$
$ a = \frac{36-4s}{3}$  -- ( 1 )
 
Total area $A = s^2 + \frac{\sqrt{3}}{4}a^2$
Substituting for $a$ from (1) above,
 
$A = s^2 + \frac{\sqrt{3}}{4} ( \frac{36-4s}{3} )^2 $
$A = s^2 + \frac{4\sqrt{3}}{9} (9-s)^2 $
 
$\frac{dA}{ds} = 2s + \frac{4\sqrt{3}}{9}(2s-18)$
$\frac{{d^2}A}{ds^2} = 2 + \frac{8\sqrt{3}}{9} > 0 $
 
Hence based on second derivative test, a minimum happens for A wherever $\frac{dA}{ds} = 0 $
 
So,
 
$\frac{dA}{ds} = 2s + \frac{4\sqrt{3}}{9}(2s-18) = 0$
 
$s = \frac{8\sqrt{3}}{2 + \frac{8\sqrt{3}}{9}} = \frac{36\sqrt{3}}{9 + 4\sqrt{3}}$
 
Substituting in ( 1 ) and simplifying, we get
 
$a = \frac{108}{9 + 4\sqrt{3}}$
 
So size of two pieces are $4s$ and $3a$ with $s$ and $a$ with values as above.
 
answered Jul 28 by pady_1
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X