Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the rectangle that will produce the largest area of the window.

Can you answer this question?

1 Answer

0 votes
Step 1:
Perimeter of the window is $P=3x+2y=12$
Area of the window =area of the rectangle+area of the triangle
$A=xy+\large\frac{\sqrt 3}{4}$$x^2$
$\;\;=x(\large\frac{12-3x}{2})+\frac{\sqrt 3}{4}$$x^2$
$\;\;=\large\frac{12x}{2}-\frac{3x^2}{2}+\frac{\sqrt 3}{4}$$x^2$
$A=6x-\large\frac{3}{2}$$x^2+\large\frac{\sqrt 3}{4}$$x^2$
Step 2:
$\large\frac{dA}{dx}$$=6-\large\frac{6x}{2}+\frac{\sqrt 3\times 2x}{4}$
$\quad=6-3x+\large\frac{\sqrt 3}{2}$$x$
For the area to be maximum or minimum
$x(\large\frac{\sqrt 3}{2}$$-3)+6=0$
$\Rightarrow x(3-\large\frac{\sqrt 3}{2})=$$6$
$x=\large\frac{6\times 2}{6-\sqrt 3}=\frac{12}{6-\sqrt 3}$
$x=\large\frac{12(6+\sqrt 3)}{3}$
$x=4(6+\sqrt 3)$
Step 3:
Diff again w.r.t $x$ we get
$\large\frac{d^2A}{dx^2}=$$-3+\large\frac{\sqrt 3}{2}$$< 0$
Hence the dimension of the rectangle produce maximum area when
$x=4(6+\sqrt 3)$ and $y=12-\large\frac{3(6+\sqrt 3)}{2}$
$y=\large\frac{24-18-3\sqrt 3}{2}$
$\;\;=6-3\sqrt 3$
$y=3(2-\sqrt 3)$
Hence $x=4(6+\sqrt 3)$ and $y=3(2-\sqrt 3)$
answered Nov 11, 2013 by sreemathi.v

Related questions

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