Answer:
equilibrium point (10,340)
Step-by-step explanation:
To find the equilibrium point, equal the demand and the supply:
[tex]D(q)=S(q)\\\\-6.10q^2-5q+1000=3.2q^2+10q-80[/tex]
Reorganize the terms in one side and reduce similar terms:
[tex]3.2q^2+6.1q^2+5q+10q-80-1000=0\\\\9.3q^2+15q-1080=0[/tex]
that's a cuadratic equation, solve with the general formula when:
a=9.3, b=15, c=-1080
[tex]q_{1}=\frac{-b+\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{1}=\frac{-15+\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{1}=\frac{-15+201}{18.6}\\\\q_{1}=\frac{186}{18.6}\\\\q_1=10[/tex]
q can't be negative because it is the quantity of bike frames, so:
[tex]q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-15-\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{2}=\frac{-15-201}{18.6}\\\\q_{2}=\frac{-216}{18.6}\\\\[/tex]
This value of q can't be considered.
Then substitute the value of q in D(q) to find the price p:
[tex]D(10) = -6.10(10)^2-5(10) + 1000\\\\D(10)=340=p[/tex]
The equilibrium point (q,p) is (10,340).
