Question

the sum of two numbers is 4. find the minimum value of the sum of their cubes.

Answer 1

Solution

Let the two numbers are $x$ and $y$

Then, according to the given condition

$x+y=4$ ... (1)

And the sum of their cubes is

$S=x^3+y^3$ ... (2)

From (1) we can write $y=4-x$, replace in (2)

$S=x^3+(4-x)^3$

We need to find the minimum sum, therefore, we minimize it using the first derivative

$\frac{dS}{dx}=3x^2+3(4-x)^2(-1)$

Replacing this first derivative equal to zero,

$\frac{dS}{dx}=0$

$3x^2+3(4-x)^2(-1)=0$

$3x^2-3(16-8x+x^2)=0$

$3x^2-48+24x-3x^2=0$

$-48+24x=0$

$x=\frac{48}{24}=2$

Which is the first number for which the sum is minimum

And therefore, the second number is

$y=4-x$

$y=4-2=2$

Hence the two numbers are $2$ and $2$