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 = 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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