Question

20 is divided into two parts so that the product of the cube of one part and the square of the other part is maximum, then these two parts are

Options

• 15, 5

• 16, 4

• 12, 8

• 14, 6

Answer 1

12, 8

Explanation:

\[\mathrm{Let}x+y=20\Rightarrow y=20-x\quad\ldots.(\mathbf{i})\] and x³y² = z

\[\Rightarrow z=x^3(20-x)^2\Rightarrow z=400x^3+x^5-40x^4\]

\[\therefore\quad\frac{\mathrm{dz}}{\mathrm{dx}}=1200x^2+5x^4-160x^3\]

For maximum or minimum,

\[\frac{\mathrm{dz}}{\mathrm{dx}}=0\]

\[\Rightarrow1200x^2+5x^4-160x^3=0\]

⇒ x = 12, 20

\[\frac{\mathrm{d}^2\mathrm{z}}{\mathrm{d}x^2}=2400x+20x^3-480x^2\]

\[\therefore\quad\left(\frac{\mathrm{d}^2\mathrm{z}}{\mathrm{d}x^2}\right)_{x=12}=-5760<0\]

∴ z is maximum at x = 12.

From (i), y = 20 − 12 = 8 

∴ The parts of 20 are 12 and 8.