Chapter 7, Problem 37RE

Solution

To calculate: The equilibrium point for the demand and supply functions.

  $\begin{array}{l}p=100-x^2\text{ Demand curve}\\ p=20+3x\text{ Supply curve}\end{array}$

The obtained result is $x=7.57,\text{ and }p=42.71$ .

Given Information:

The function of demand and supply are defined as,

  $\begin{array}{l}p=100-x^2\text{ Demand curve}\\ p=20+3x\text{ Supply curve}\end{array}$

Calculation:

Consider the given functions,

  $\begin{array}{l}p=100-x^2\text{ Demand curve}\\ p=20+3x\text{ Supply curve}\end{array}$

Equate both the function equal to find the equilibrium point.

  $\begin{array}{c}20+3x=100-x^2\\ x^2+3x+20-100=0\\ x^2+3x-80=0\end{array}$

Use the quadratic formula to find the solution.

  $\begin{array}{c}x=\frac{-3\pm \sqrt{3^2-4\left(-80\right)}}{2}\\ =\frac{-3\pm 18.14}{2}\\ =7.57,-10.57\end{array}$

The value cannot be negative. So take the positive value.

Substitute the value in the supply function and calculate p .

  $\begin{array}{c}p=20+3\left(7.57\right)\\ =42.71\end{array}$

Hence, the obtained values are $x=7.57,\text{ and }p=42.71$ .