Question 4: A company’s profit \(\small{P(x)}\) in rupees from selling \(\small{x}\) units of a product is modeled by the cubic expression:
\[P(x)=x^3-15x^2+75x-125\]
Find the break-even point(s), where the profit is zero.
Solution:
$P(x)=x^{3} -15x^{2} +75x-125$
$\ \ \ \ \ \ \ =x^{3} -3\left( x^{2}\right)( 5) +3( x)\left( 5^{2}\right) -5^{3}$
$\ \ \ \ \ \ \ =( x-5)^{3}$
For zero profit $\displaystyle P( x) =0$
$\displaystyle ( x-5)^{3} =0$
$\displaystyle x-5=0$
$\displaystyle x=5$
The break even point is $\displaystyle 5$, where profit is zero.