Question 1129290

 if {{{C = 40x + 800 }}} and the revenue {{{R = 100x -0.5x^2}}}

profit is {{{P=R-C}}}

{{{P=100x -0.5x^2-(40x + 800)}}}

{{{P=100x -0.5x^2-40x - 800}}}

{{{P=-0.5x^2+60x - 800}}}

profit of {{{P=800}}}


{{{800=-0.5x^2+60x - 800}}}


{{{0=-0.5x^2+60x - 800-800}}}

{{{-0.5x^2+60x - 1600=0}}}...factor

{{{-0.5(x^2-120x + 3200)=0}}}


write {{{-120x}}} as {{{-40x-80x}}}, than we have


{{{-0.5(x^2-40x -80x+3200)=0}}}


{{{-0.5((x^2-80x) -(40x- 3200))=0}}}


{{{-0.5(x(x-80) -40(x- 80))=0}}}


{{{-0.5 (x - 80) (x - 40) = 0}}}

solutions:

{{{ (x - 80) = 0}}}=>{{{x=80}}}

{{{ (x - 40) = 0}}}=>{{{x=40}}}