Divide 15 into two parts such that the square of one number multiplied with the cube of the other number is maximum.

Given,

• the number 15 is divided into two numbers.

• the product of the square of one number and cube of another number is maximum.

Let us consider,

• x and y are the two numbers

• Sum of the numbers : x + y = 15

• Product of square of the one number and cube of anther number : P = x³ y²

Now as,

x + y = 15

y = (15-x) ------ (1)

Consider,

P = x³y²

By substituting (1), we have

P = x³ × (15-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 and if u and v are two functions of x, then ]

= 3×[15² – 2× (15)×(x) + x²] x² + x³(2x-30)

= x²[3× (225 – 30x + x²)+ x (2x - 30)]

= x²[ 675– 90x + 3x²+ 2x² – 60x]

= x²[5x² – 120x + 675]

= 5x² [x² – 24x + 135] ----- (3)

Now equating the first derivative to zero will give the critical point c.

So,

Hence 5x² = 0 (or) x² – 24x + 135 = 0

x = 0 (or)

x = 0 (or)

x = 0 (or)

x = 0 (or)

x = 0 (or) (or)

x = 0 (or) (or)

x = 0 (or) x = 15 (or) x = 9

Now considering the critical values of x = 0,9,15

Now, for checking if the value of P is maximum or minimum at x=0,9,15, we will perform the second differentiation and check the value of at the critical value x = 0,9,15.

Performing the second differentiation on the equation (3) with respect to x.

= (x² – 24x + 135) (5 × 2x) + 5x² (2x – 24 + 0)

[Since and and if u and v are two functions of x, then ]

= (x² – 24x + 135) (10x) + 5x² (2x – 24)

= 10x³ – 240x² + 1350x + 10x³ – 120x²

= 20x³ – 360x² + 1350x

= 5x (4x² – 72x + 270)

Now when x = 0,

= 0

So, we reject x = 0

Now when x = 15,

= 65 [(4 × 225) –1080+ 270]

= 65 [900– 1080+ 270]

= 65 [1170– 1080]

= 65× (90) > 0

Hence , so at x = 15, the function P is minimum

Now when x = 9,

= 45 [(4 × 81) – 648 + 270]

= 45 [324 – 648 + 270]

= 45 [594 – 648]

= 45 × (-54)

= -2430 < 0

As second differential is negative, hence at the critical point x = 9 will be the maximum point of the function P.

Therefore, the function P is maximum at x = 9.

From Equation (1), if x= 9

y = 15 – 9 = 6

Therefore, x = 9 and y = 6 are the two positive numbers whose sum is 15 and the product of the square of one number and cube of another number is maximum.