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.