CBSE Class 12-science Answered

A window is in the form of a rectangle surmounted by a semi circular opening. The total perimeter of the window is 10 m. Find the dimensions o the window to admit maximum light through the whole opening. Do you think that by getting maximum light, we can save electricity? Do you agree that we should save electricity?

Asked by Topperlearning User | 19 Aug, 2014, 09:38: AM

Expert Answer

$ATQ space the space shape space of space the space window space will space be space as space follows : let space us space take space the space window space of space length space 2 straight x space units space so space the space radius space of space semi minus circular space opening space would space be space straight x space units. space Taking space height space of space the space rectangular space portion space of space window to space be space straight y space units comma space we space get space he space length space of space boundary space as space 10 equals 2 straight x plus 2 straight y plus πx space donot space include space the space line space segment space AS space as space this space is space not space the space part space of space window. Area space of space the space window space is space given space as straight A equals 2 xy plus πx squared over 2 After space replacing space straight y space by space fraction numerator 10 minus 2 straight x minus πx over denominator 2 end fraction we space get space straight A equals 2 straight x open parentheses fraction numerator 10 minus 2 straight x minus πx over denominator 2 end fraction close parentheses plus πx squared over 2 rightwards double arrow straight A equals 10 straight x minus 2 straight x squared minus πx squared plus πx squared over 2 rightwards double arrow straight A equals 10 straight x minus 2 straight x squared minus πx squared over 2 Differentiating space straight w. straight r. straight t. space straight x space dA over dx equals 10 minus 4 straight x minus πx Taking space dA over dx equals 0 rightwards double arrow 10 minus 4 straight x minus πx equals 0 rightwards double arrow straight x equals fraction numerator 10 over denominator straight pi plus 4 end fraction To space check space for space maxima space or space minima space fraction numerator straight d squared straight A over denominator dx squared end fraction equals minus 4 minus straight pi space which space is space surely space negative space so space it gives space maximum space area space when space straight x equals fraction numerator 10 over denominator straight pi plus 4 end fraction space and space correspondingly space straight y equals fraction numerator 10 over denominator straight pi plus 4 end fraction. So space the space length space be space fraction numerator 10 over denominator straight pi plus 4 end fraction of space the space rectangle space portion space to space admit space maximum space sunlight. Yes comma space by space getting space maximum space light comma space we space can space save space electricity. space We space must space save space electricity as space and space when space possible.$