Browse Questions

# Find two positive numbers x and y such that their sum is 16 and sum of whose cubes is minimum.

Toolbox:
• For greatest area, maximise area.
• Working rule for maxima minima.
• Find the function to be maximised or minimised
• Convert the function into single variable
• Apply the conditions for maximum or minimum i.e., first derivative = 0 ( w.r.to the variable converted) and second derivative positive for minima and negative for maxima.
Step 1
Given $x+y=16$
and $s=x^3+y^3$ is minimum.
$s = x^3+(16-x)^3$
Step 2
$\large\frac{ds}{dx}=3x^2-3(16-x)^2=0$
$\Rightarrow x = 8\: \: y = 8$
Step 3
$\large\frac{d^2s}{dx^2}=6x+6(16-x)$ is positive at $x = 8$
$\Rightarrow s$ is minimum.
Ans : $x = 8, \: \: y = 8$