Divide a into two parts such that the product of the pth power of one part and the qth power of the second part may be maximum.

Given,


the number ‘a’ is divided into two numbers.


the product of the pth power of one number and qth power of another number is maximum.


Let us consider,


x and y are the two numbers


Sum of the numbers : x + y = a


Product of square of the one number and cube of anther number : P = xp yq


Now as,


x + y = a


y = (a-x) ------ (1)


Consider,


P = xpyq


By substituting (1), we have


P = xp × (a-x)q ------ (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 ]



= xp-1(a-x)q-1[ap-xp-xq]


= xp-1(a-x)q-1[ap – x (p+q)] ----- (3)


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


So,



Hence xp-1 = 0 (or) (a-x)q-1 (or) ap– x(p+q)= 0


x = 0 (or) x = a (or)


Now considering the critical values of x = 0,a and


Now, using the First Derivative test,


For f, a continuous function which has a critical point c, then, function has the local maximum at c, if f’(x) changes the sign from positive to negative as x increases through c, i.e. f’(x)>0 at every point close to the left of c and f’(x)<0 at every point close to the right of c.


Now when x = 0,



So, we reject x = 0


Now when x = a,



Hence we reject x = a


Now when ,


> 0 ---- (4)


Now when ,


< 0 ---- (5)


By using first derivative test, from (4) and (5), we can conclude that, the function P has local maximum at


From Equation (1), if



Therefore, and are the two positive numbers whose sum together to give the number ‘a’ and whose product of the pth power of one number and qth power of the other number is maximum.


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