# Find the equilibrium price given demand and supply equations: D = -5p + 40, S = -p^2 + 30p - 8.

## Question:

Find the equilibrium price given demand and supply equations:

{eq}D = -5p + 40 {/eq}

{eq}S = -p^2 + 30p - 8 {/eq}

## Quadratic Formula:

An equation which is of the form {eq}ax^2+bx+c=0 {/eq} where {eq}a, b \text{ and } c {/eq} are constants is called a quadratic equation in {eq}x {/eq}. It has two (either same or distinct) roots. The roots can be solved by using the quadratic formula which states:

$$x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} $$

## Answer and Explanation:

The demand and supply functions are:

$$\begin{align} D &= -5p + 40\\[0.5cm] S& = -p^2 + 30p -8 \end{align} $$

To find the equilibrium price, we set the supply equal to demand and solve for {eq}p {/eq}:

$$D=S \\[0.5cm] -5p+40=-p^2+30p-8 \\[0.5cm] \text{Adding }p^2 \text{ and 8 and subtracting 30 p from both sides}, \\[0.5cm] p^2 -35p +48=0 $$

Comparing this with {eq}ap^2+bp+c=0 {/eq}, we get:

$$a=1\\[0.5cm] b=-35\\[0.5cm] c=48 $$

Substitute these values in the quadratic formula:

$$\begin{align} p&=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\[0.5cm] &=\frac{-(-35) \pm \sqrt{(-35)^{2}-4 \cdot 1 \cdot 48}}{2 \cdot 1} \\[0.5cm] &= \frac{35\pm \sqrt{1033}}{2}\\[0.5cm] & = \frac{35+\sqrt{1033}}{2}; \,\,\, \frac{35-\sqrt{1033}}{2} \\[0.5cm] & \approx 33.57; \,\,\,\,\,\,\,\,\,1.43 & [ \text{The answers are rounded to 2 decimals} ] \end{align} $$

But if we substitute {eq}p=33.57 {/eq} in the demand function, we get {eq}D = -5(33.57) +40 = -127.85 {/eq}, a negative value.

So the value of {eq}p {/eq} cannot be {eq}33.57 {/eq}.

**Therefore, the equilibrium price = {eq}\color{blue}{\boxed{\mathbf{\$ 1.43}}} {/eq}.**

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from Math 101: College Algebra

Chapter 4 / Lesson 10