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# Find the two numbers whose sum is 15 and the square of onemultiplied by the cube of the other is maximum.

290 Points
9 years ago

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,

Swapnil Saxena
102 Points
9 years ago

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 -30y4 )

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