Find the two numbers whose sum is 15 and the square of one

multiplied by the cube of the other is maximum.

Hi Shaleen.

Lets consider the two numbers to be x,y (both greater than 0).

So x+y = 15

Now to maximise T = x^2 * y^3

So T = y^3 (15-y)^2 = y^3 { 225 + y^2 - 30y)

So T = 225y^3 + y^5 - 30y^4.

Now maximise this using dT/dy = 0.

Clearly y=0 will not maximise "T".

Hence y(2y-30) + 3 { 225 + y^2 - 30y } = 0.

Solve this QE, for y value (for which the maximum occurs).

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

Algorithm:

First form a equations i. x+y=15 ----(1) and (x^2)(y^3) is maximum

Now take the value of x from equation  (1).

=((15-y)²)(y³)

=(225+y²-30y)(y³)

=(225y³ +y⁵ -30y^{4 )}

Now differentiate in terms of y and putting it equal to 0 which is the slope of the graph at the maxima or minima

0=675y²+5y⁴-120y³

0=5y²(y²-24y+135)

y2-24y+135=0

The above equation are satisfied only at 15,9

At 15,0 it is at the minima, So y=9 and x=6 must be the correct answers