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 $P(p)=-5p^2+120p-540:$

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

Hence break-even points occur when $p=6$ and $p=18.$ 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 $q=-5(6-16)+20=70$ and $q=-5(18-16)+20=10.$