Science:Math Exam Resources/Courses/MATH104/December 2015/Question 08 (c)/Solution 1

Continuing from part (b), we find the roots of the profit function $P(p)=-5p^{2}+120p-540:$

${\begin{aligned}0&=-5p^{2}+120p-540\\&=-p^{2}+24p-108\\&=-p^{2}+6p+18p-108\\&=-p(p-6)+18(p-6)\\&=(18-p)(p-6).\end{aligned}}$

Hence break-even points occur when $p={\color {OliveGreen}6}$ and $p={\color {PineGreen}18}.$ Typically, break-even points are expressed as demand ($q$) values for which the profit is zero, so we use our result $q(p)=-5(p-16)+20$ from part (a) to convert these to $q=-5({\color {OliveGreen}6}-16)+20={\color {blue}70}$ and $q=-5({\color {PineGreen}18}-16)+20={\color {blue}10}.$