Question

The owner of a bike shop that produces custom built bike frames has determined that the demand equation for bike frames is given by the equation
D(q) = –6.10q^2 –5q + 1000
where D(q) is the price in dollars and q is the number of bike frames demanded per week. The supply equation for bike frames is
S(q) = 3.20q^2 + 10q – 80
where q is the quantity the supplier will make available per week in the market when the price is p dollars. Find the equilibrium point (q, p) rounded to the nearest hundredth.

Answer 1

Answer:

equilibrium point (10,340)

Step-by-step explanation:

To find the equilibrium point, equal the demand and the supply:

$D(q)=S(q)\\\\-6.10q^2-5q+1000=3.2q^2+10q-80$

Reorganize the terms in one side and reduce similar terms:

$3.2q^2+6.1q^2+5q+10q-80-1000=0\\\\9.3q^2+15q-1080=0$

that's a cuadratic equation, solve with the general formula when:

a=9.3, b=15, c=-1080

$q_{1}=\frac{-b+\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{1}=\frac{-15+\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{1}=\frac{-15+201}{18.6}\\\\q_{1}=\frac{186}{18.6}\\\\q_1=10$

q can't be negative because it is the quantity of bike frames, so:

$q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-15-\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{2}=\frac{-15-201}{18.6}\\\\q_{2}=\frac{-216}{18.6}$

This value of q can't be considered.

Then substitute the value of q in D(q) to find the price p:

$D(10) = -6.10(10)^2-5(10) + 1000\\\\D(10)=340=p$

The equilibrium point (q,p) is (10,340).