Question

Find two numbers whose sum is 15 and when the square of one number multiplied by the cube of the other is maximum.

Answer 1

Let the two numbers be x and y.

Then x + y = 15 

∴ y = 15 − x

Let P is the product of square of y and cube of x.

Then P = x³y²

∴ P = x³(15 − x)²

∴ P = x³(225 − 30x + x²)

∴ P = x⁵ − 30x⁴ + 225x³

∴  `(dP)/(dx) = (d)/(dx) (x^5 − 30x^4 + 225x^3)`

∴  `(dP)/(dx) = 5x^4 − 30 × 4x^3 + 225 × 3x^2`

∴  `(dp)/(dx) = 5x^4 − 120x^3 + 675x^2`

and `(d^2P)/(dx^2) = (d)/(dx)(5x^4 - 120x^3 + 675x^2)`

`(d^2P)/(dx^2) = 5 xx 4x^3 - 120 xx 3x^2 + 675 xx 2x`

`(d^2P)/(dx^2) = 20x^3 - 360x^2 + 1350x`

`(d^2P)/(dx^2) = 10x(2x^2 - 36x + 135)`

Now, `(dP)/(dx)` = 0 gives 5x⁴ − 120x³ + 675x² = 0

∴ 5x²(x² − 24x + 135) = 0

∴ 5x²(x² − 15x − 9x + 135) = 0

∴ 5x²[x(x − 15) − 9(x − 15)] = 0

∴ 5x²(x − 15)(x − 9) = 0

∴ The roots of `(dP)/(dx)` = 0 are x₁ = 0, x₂ = 15 and x₃ = 9

If x = 0, then y = 15 − 0 = 15

If x = 15, then y = 15 − 15 = 0

In both the cases product x³y² are zero, which is not maximum.

∴ x ≠ 0 and x ≠ 15          

∴ x = 6

Now, `((d^2P)/(dx^2))_("at x = 6")` = 10(6)[2(6)² − 36 × 6 + 135]

= 60[72 − 216 + 135]

= 60(− 9)

= − 540 < 0

∴ P is maximum when x = 6

If x = 6, then y = 15 − 6 = 9

Hence, the required numbers are 6 and 9