Answer :

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 = x^{3} y^{2}

Now as,

x + y = 15

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

Consider,

P = x^{3}y^{2}

By substituting (1), we have

P = x^{3} × (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×[15^{2} – 2× (15)×(x) + x^{2}] x^{2} + x^{3}(2x-30)

= x^{2}[3× (225 – 30x + x^{2})+ x (2x - 30)]

= x^{2}[ 675– 90x + 3x^{2}+ 2x^{2} – 60x]

= x^{2}[5x^{2} – 120x + 675]

= 5x^{2} [x^{2} – 24x + 135] ----- (3)

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

So,

Hence 5x^{2} = 0 (or) x^{2} – 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.

= (x^{2} – 24x + 135) (5 × 2x) + 5x^{2} (2x – 24 + 0)

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

= (x^{2} – 24x + 135) (10x) + 5x^{2} (2x – 24)

= 10x^{3} – 240x^{2} + 1350x + 10x^{3} – 120x^{2}

= 20x^{3} – 360x^{2} + 1350x

= 5x (4x^{2} – 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.

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