Application of Derivatives

Question

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 of the window to admit maximum light through the whole opening.

Solution

Let x, y be length, breadth of the rectangle ABCD.
Let

straight x over 2

be the radius of the semi-circle with centre at O.
Perimeter of figure = 10 meters

therefore space space space space straight x plus 2 straight y plus straight pi straight x over 2 space equals space 10 space space space space space space space rightwards double arrow space space 2 straight x plus 4 straight y plus πx space equals 20 therefore space space space space space space 4 straight y space equals 20 minus left parenthesis straight pi plus 2 right parenthesis straight x rightwards double arrow space space space space space straight y space equals space fraction numerator 20 minus left parenthesis straight pi plus 2 right parenthesis space straight x over denominator 4 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space space space space space

Let A be the area of the figure.

therefore space space space straight A space equals space xy space plus 1 half straight pi open parentheses straight x over 2 close parentheses squared space equals straight x open square brackets fraction numerator 20 minus left parenthesis straight pi plus 2 right parenthesis space straight x over denominator 4 end fraction close square brackets plus πx squared over 8

open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets

therefore space space space space straight A space equals space left square bracket 20 straight x minus left parenthesis straight pi plus 2 right parenthesis straight x squared right square bracket space plus πx squared over 8 therefore space space space dA over dx space equals 1 fourth left square bracket 20 minus space 2 left parenthesis straight pi plus 2 right parenthesis straight x right square bracket plus space πx over 4 Now space space dA over dx equals 0 space space space space space rightwards double arrow space space space space 1 fourth left square bracket 20 minus 2 left parenthesis straight pi minus 2 right parenthesis straight x right square bracket plus πx over 4 equals 0 rightwards double arrow space space space space space 20 minus space 2 left parenthesis straight pi plus 2 right parenthesis space straight x space plus space πx space equals space 0 rightwards double arrow space space space space space space 20 minus 2 πx minus 4 straight x plus πx space equals space 0 rightwards double arrow space space space space space 20 minus left parenthesis straight pi plus 4 right parenthesis straight x space equals space 0 space space space space rightwards double arrow space space space straight x space equals space fraction numerator 20 over denominator straight pi plus 4 end fraction

fraction numerator straight d squared straight A over denominator dx squared end fraction space equals space 1 fourth left square bracket 0 minus 2 left parenthesis straight pi plus 2 right parenthesis right square bracket space plus straight pi over 4 space equals 1 fourth left square bracket negative 2 straight pi minus 4 plus straight pi right square bracket space equals negative fraction numerator straight pi plus 4 over denominator 4 end fraction

When space straight x space equals fraction numerator 20 over denominator straight pi plus 4 end fraction. space fraction numerator straight d squared straight A over denominator dx squared end fraction space equals negative fraction numerator straight pi plus 4 over denominator 4 end fraction less than 0 therefore space space space space space space space space space space straight A space is space maximum space when space straight x space equals fraction numerator 20 over denominator straight pi plus 4 end fraction

and

straight y space equals space fraction numerator 20 minus left parenthesis straight pi plus 2 right parenthesis space begin display style fraction numerator 20 over denominator straight pi plus 4 end fraction end style over denominator 4 end fraction space equals space fraction numerator 20 straight pi plus 80 minus 20 straight pi minus 40 over denominator 4 left parenthesis straight pi plus 4 right parenthesis end fraction space equals space fraction numerator 10 over denominator straight pi plus 4 end fraction

therefore

area of figure is maximum i.e., maximum light is admitted when length of rectangle =

straight x equals fraction numerator 20 over denominator straight pi plus 4 end fraction comma space space breadth space of space rectangle space equals straight y space equals space fraction numerator 10 over denominator straight pi plus 4 end fraction comma

radius space of space semi minus circle space equals space straight x over 2 space equals space fraction numerator 5 over denominator straight pi plus 4 end fraction.