Question 1124409

{{{C = 50x + 600}}} and the revenue is given by {{{R = 100x -0.5x^2}}}.

 Recall that profit is revenue minus cost.

{{{P=R-C}}}

{{{P=100x -0.5x^2-(50x + 600)}}}

{{{P=100x -0.5x^2-50x - 600}}}

{{{P=-0.5x^2+50x  - 600}}}


find two values of {{{x }}}(production level) that will create a profit of ${{{600}}}:


{{{600=-0.5x^2+50x  - 600}}}

{{{0.5x^2-50x+600+600=0 }}}

{{{0.5x^2-50x+1200=0 }}}

{{{0.5(x^2-100x+2400)=0 }}}

{{{0.5(x^2-40x-60x+2400)=0 }}}

{{{0.5((x^2-40x)-(60x-2400))=0 }}}

{{{0.5(x(x-40)-60(x-40))=0 }}}

{{{0.5 (x - 60) (x - 40) = 0 }}}

since {{{0.5<>0}}}, equation will be zero if {{{(x - 60)  = 0 }}} or {{{  (x - 40) = 0 }}}

if {{{(x - 60)  = 0 }}}->{{{x=60}}}
if {{{  (x - 40) = 0 }}}->{{{x=40}}}

so, the  values of x (production level) that will create a profit of ${{{600}}} are {{{x=40}}} and {{{x=60}}}