1. Class 12
2. Important Question for exams Class 12
3. Chapter 6 Class 12 Application of Derivatives

Transcript

Misc 11 A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. Let Length of rectangle be x & breadth of rectangle be y Diameter of semicircle = x Radius of semicircle = 2 Given , Perimeter of window = 10 m Length + 2 Breadth + circumference of semicircle = 10 +2 + 2 =10 2 =10 2 =5 2 2 =5 1 2 + 4 We need to maximize area of window Area of window = Area of rectangle + Area of Semicircle A = + 1 2 2 2 A = 5 1 2 + 4 + 2 8 A = 5 5 2 2 2 4 + 2 8 A = 5 5 2 2 2 8 Finding = 5 5 2 2 2 8 =5 5 4 Putting =0 0 = 5 5 4 5 4 =5 = 5 5 4 = 20 + 4 Now, calculating 2 2 at = 20 + 4 2 2 = 5 5 4 2 2 = 5 4 <0 So, 2 2 <0 at = 20 + 4 Hence, = 20 + 4 maxima Hence, A is maximum when = 20 + 4 Now, Finding value of y =5 1 2 + 4 =5 20 + 4 1 2 + 4 =5 20 + 4 4 + 2 2 =5 5 2 4 + 2 + 4 =5 5 2 + + 4 =5 + 4 2 + + 4 = 5 2 + 4 = 10 + 4 Hence, for maximum area, Length = = + m & Breadth = = + m

Chapter 6 Class 12 Application of Derivatives

Class 12
Important Question for exams Class 12