Divide 8 into two positive parts such that the sum of the square of one and the cube of the other is minimum.
Given,
• the number 8 is divided into two numbers.
• the product of the square of one number and cube of another number is minimum.
Let us consider,
• x and y are the two numbers
• Sum of the numbers : x + y = 8
• Product of square of the one number and cube of anther number : S = x3 + y2
Now as,
x + y = 8
y = (8-x) ------ (1)
Consider,
S = x3 + y2
By substituting (1), we have
S = x3 + (8-x)2 ------ (2)
For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.
Differentiating the equation (2) with x
[Since 
]
= 3x2 + 2x – 16 ------ (3)
Now equating the first derivative to zero will give the critical point c.
So,
Hence 3x2 + 2x - 16= 0
(or) 
(or) 
x = 2 (or) x = -2.67
Now considering the critical values of x = 2,-2.67
Now, for checking if the value of P is maximum or minimum at x=2,-2.67, we will perform the second differentiation and check the value of 
at the critical value x = 2,-2.67.
Performing the second differentiation on the equation (3) with respect to x.
= 3 (2x) + 2 (1) - 0
[Since 
and 
]
= 6x + 2
Now when x = -2.67,
= - 16.02 + 2 = -14.02
At x = -2.67 
hence, the function S will be maximum at this point.
Now consider x = 2,
= 12 + 2 = 14
Hence 
, so at x = 2, the function S is minimum
As second differential is positive, hence at the critical point x = 2 will be the maximum point of the function S.
Therefore, the function S is maximum at x = 2.
From Equation (1), if x= 2
y = 8 – 2 = 6
Therefore, x = 2 and y = 6 are the two positive numbers whose sum is 8 and the sum of the square of one number and cube of another number is maximum.