Question
Sun March 15, 2015 By: Ravi Consul

Q:A window is in the form of a rectangle surmounted by a semicircle.If the total perimeter of the window is 30m ,find the dimensions of the window so that maximum light is allowed. 

Expert Reply
Vijaykumar Wani
Mon March 16, 2015
begin mathsize 14px style The space perimeter space is space the space lengths space of space three space sides space of space the space rectangle space plus space half space the space circumference space of space straight a space circle space of space radius space straight r. Also space the space area space is space the space area space of space the space rectangle space plus space half space the space area space of space straight a space circle space of space radius space straight r. straight A equals 2 hr plus 1 half πr squared........ left parenthesis 1 right parenthesis The space perimetre space of space window space is space 30 straight m 30 equals 2 straight h plus 2 straight r plus πr rightwards double arrow straight h equals 15 minus straight r minus πr over 2.... left parenthesis 2 right parenthesis   Put space value space of space straight h space from space eq space left parenthesis 2 right parenthesis space in space eq space left parenthesis 1 right parenthesis space we space get straight A equals 2 open parentheses 15 minus straight r minus πr over 2 close parentheses straight r plus 1 half πr squared straight A equals 30 straight r minus 2 straight r squared minus πr squared plus 1 half πr squared straight A equals 30 straight r minus 2 straight r squared minus 1 half πr squared  Differentiate space straight A space straight w. straight r. straight t. space straight r space we space get comma straight A apostrophe space equals 30 minus 4 straight r minus πr  Differentiate space straight A apostrophe space straight w. straight r. straight t. space straight r space we space get comma straight A apostrophe apostrophe space equals negative 4 minus straight pi  We space can space see space that space the space only space critical space point space is comma straight r equals fraction numerator 30 over denominator 4 plus straight pi end fraction equals 4.2016 We space can space also space see space that space the space second space derivative space is space always space negative comma space So space maximum space area space must space occour space at space this space point. Put space straight r equals 4.2016 space in space eq space left parenthesis 2 right parenthesis space we space get straight h equals 4.20188  So space the space dimensins space are open parentheses straight h cross times 2 straight r close parentheses equals 4.20188 cross times 4.2016  end style
 
 
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