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Show that of all the rectangles in a given fixed circle, the square has the maximum area. ` (##NTN_MATH_XII_C06_E04_252_Q01.png" width="80%"> |
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Answer» Let r be the radius of given circle. A rectangle ABCD is inscribed in the circle whose diagonal is AC= 2r. Let ` AB = x and BC = y` `:." In "Delta ABC`, `AB^(2)+BC^(2)=AC^(2)` ` rArr x^(2) + y^(2) = (2r)^(2)` ` rArr y^(2) = 4r^(2) - x^(2)` ...(1) Now, area of rectangle `A = x * y` ` rArr A = x sqrt(4r^(2)-x^(2))` ` rArr A^(2) = x^(2) (4r^(2)-x^(2))` Let ` Z=A^(2)` ` rArr Z = 4r^(2)x^(2) - x^(4)` ` rArr (dz)/(dx) = 8r^(2)x-4x^(3)` For maxima/minima ` (dZ)/(dx) = 0` ` rArr 8r^(2)x-4x^(3) = 0 rArr x^(2)=2r^(2)` and ` ((d^(2)Z)/(dr^(2))) = 8r^(2) - 12x^(2)` at ` x^(2) =2r^(2)` ` (d^(2)Z)/(dr^(2)) = 8r^(2) - 24r^(2) =- 16r^(2) lt 0 ` ` rArr at x^(2) = 2r^(2) , Z` is maximum. ` rArr at x^(2) = 2r^(2), A^(2)` is maximum. ` rArr at x^(2) = 2r^(2), A` is maximum. Now ` x^(2) = 2r^(2)` ` rArr 2x^(2) = 4r^(2) = x^(2)+y^(2)` [From equation (1)] ` rArr x^(2) = y^(2)` ` rArr x = y ` Therefore, the rectangle with maximum area, is a square. |
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