The number 28 is divided into two positive parts such that the sum of the cube of one part and the square of the other part is minimum, then the absolute difference between the two parts is

Step-by-step solution

Let the two positive parts of 28 be x and 28 - x , where 0 Let S be the sum of the cube of one part and the square of the other part. Then, S = x^3 + (28 - x)^2 To find the minimum value of S , we differentiate S with respect to x and equate it to zero: dS dx = 3x^2 + 2(28 - x)(-1) dS dx = 3x^2 + 2x - 56 Setting dS dx = 0 , we get: 3x^2 + 2x - 56 = 0 3x^2 + 14x - 12x - 56 = 0 x(3x + 14) - 4(3x + 14) = 0 (x - 4)(3x + 14) = 0 Since x must be positive, we have x = 4 . Now, we check the second derivative to confirm it