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how do i calculate it?

Qd = 900 - 10p

p = 15 + 0.05Qs

This is what have tried.

making Qs the subject of formular

p - 15 = 0.05Qs

Qs= (p-15)/0.05

then now to solve the equlibrium

Qd=Qs

900 - 10p = (p-15)/0.05

(900 - 10p)* 0.05 = p - 15

45 - 0.5p = p - 15

45 + 15 = p + 0.05p

60 = 1.05p

p = 60/1.05 = 57.1

insert p into Qd

Qd = 900 - 10(57.1)

   = 900 - 571

 Qd  = 329
  • 0
    Yes that's right, this is the usual approach to this kind of problems – 2017-02-25
  • 0
    if it is correct, how come when insert it into Qs i got different answer? Qs = (57.1 -15)/0.05 = 842 – 2017-02-25
  • 0
    Problem was due to a calculation error, look at my answer – 2017-02-25

1 Answers 1

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The equilibrium of price and quantity is at the point $(p^*,q^*)$ where the following equation is satisfied: $$Q_d=q^*=Q_s$$

Thus we want to find the unique solution of this system of equations: $$ \left\{ \begin{array}{c} 10p+Q_d=900 \\ p-0.05Q_s=15 \end{array} \right. $$

Firstly we set $Q_d=q^*=Q_s$, which gives us: $$ \left\{ \begin{array}{c} 10p+q^*=900 \\ p-0.05q^*=15 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p \\ p-0.05\cdot(900-10p)=15 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p \\ \frac32p=45+15=60 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p \\ p^*=p=40 \end{array} \right. $$ $$\Leftrightarrow \left\{ \begin{array}{c} q^*=900-10p^*=900-10\cdot 40=500 \\ p^*=40 \end{array} \right. $$

Thus the equilibrium point is $(p^*,q^*)=(40,500)$.

Let's control if this is the right answer:

  • $Q_d=900-10\cdot 40=500$
  • $Q_s=\frac {p-15}{0.05}=\frac {40-15}{0.05}=\frac{25}{0.05}=500$