Question

Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Answer 1

Suppose the given two numbers be x and y. Now,

x + y = 15 ….. (1)

Then we have, z = x² y³

z = x² (15 – x)³ (from the equation 1)

dz/dx = 2x(15 - x)³ - 3x²(15 - x)²

For the maximum or minimum values of z, we must have

dz/dx = 0

Hence, z is the maximum when x = 6 and y = 9

Therefore, the required two parts into which 15 should be divided are 6 and 9.