Question

The sum of two positive numbers is given. If the sum of their cubes is minimum, then

• A. they are equal
• B. one is twice the other
• C. they are unequal
• D. one is thrice the other

Answer 1

Answer: (A)

Let x and y be two positive numbers.
$\therefore$ According to question,
$x+y=a$ (constant)
$\Rightarrow y=a-x$
Let $z=x^3+y^3$
$z=x^3+{\left(a-x\right)}^3...\left(i\right)$
On differentiating Eq. (i) w.r .t. 'x', we get
$\frac{dz}{dx}=3x^2+3{\left(a-x\right)}^2\left(-1\right)$
$=3\left[x^2-{\left(a-x\right)}^2\right]$
$=3\left(x+a-x\right)\left(x-a+x\right)$
$=3a\left(2x-a\right)...\left(ii\right)$
For maximum or minimum, $\frac{dz}{dx}=0$
$\Rightarrow 3a\left(2x-a\right)=0\left(\because 3a\ne 0\right)$
$\Rightarrow 2x-a=0$
$\Rightarrow x=\frac{a}{2}$
On differentiating Eq. (ii) w.r.t 'x', we get
$\frac{d^{2}z}{d{x}^2}=6a$ (positive)
$\therefore \frac{d^{2}z}{d{x}^2}>0$
$\therefore$ z has a minimum value at $x=\frac{a}{2}$.
Therefore, the numbers are $\frac{a}{2}$ and $a-\frac{a}{2}=\frac{a}{2}.$