Question 90296
To find the equilibrium price, set D equal to S

{{{-200p+35000=-p^2 + 400p - 20000}}}


{{{-200p+35000+p^2=400p-20000}}} Add {{{p^2}}} to both sides


{{{-200p+35000+p^2-400p=-20000}}}Subtract {{{400p}}} from both sides


{{{-200p+35000+p^2-400p+20000=0}}} Add {{{20000}}} to both sides


{{{p^2-600p+55000=0}}} Combine like terms




Now let's use the quadratic formula to solve for p:



Starting with the general quadratic


{{{ap^2+bp+c=0}}}


the general solution using the quadratic equation is:


{{{p = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{p^2-600*p+55000=0}}} ( notice {{{a=1}}}, {{{b=-600}}}, and {{{c=55000}}})


{{{p = (--600 +- sqrt( (-600)^2-4*1*55000 ))/(2*1)}}} Plug in a=1, b=-600, and c=55000




{{{p = (600 +- sqrt( (-600)^2-4*1*55000 ))/(2*1)}}} Negate -600 to get 600




{{{p = (600 +- sqrt( 360000-4*1*55000 ))/(2*1)}}} Square -600 to get 360000




{{{p = (600 +- sqrt( 360000+-220000 ))/(2*1)}}} Multiply {{{-4*55000*1}}} to get {{{-220000}}}




{{{p = (600 +- sqrt( 140000 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{p = (600 +- 100*sqrt(14))/(2*1)}}} Simplify the square root




{{{p = (600 +- 100*sqrt(14))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{p = (600 + 100*sqrt(14))/2}}} or {{{p = (600 - 100*sqrt(14))/2}}}



which split up to



{{{p=+600/2+100*sqrt(14)/2}}} or {{{p=+600/2-100*sqrt(14)/2}}}



and simplify to



{{{p=300+50*sqrt(14)}}} or {{{p=300-50*sqrt(14)}}}



Which approximate to


{{{p=487.082869338697}}} or {{{p=112.917130661303}}}



So our solutions are:

{{{p=487.082869338697}}} or {{{p=112.917130661303}}}




However, since the first solution {{{x=487.082869338697}}} produces a negative demand value, we must discard the first solution.


So the equilibrium price is $112.92