Q. 3

# Divide 15 into tw

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.

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