A company wholesales shampoo in a particular city. Their marketing research department established the following weeklyrsupply-and-demand equations. \[ p=\frac{x}{600}+\frac{1}{2} \quad \text { Supply equation } \] \[ p=\frac{4,500}{x} \quad \text { Demand equation } \] How many units are required for supply to equal demand? At what price per bottle will supply equal demand? How many units are required for supply to equal demand? At what price per bottle will supply equal demand? \( \$ \square \) (Round to the nearest cent as needed.)
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To find the quantity where supply equals demand, we need to set the supply equation equal to the demand equation. The supply equation is: \[ p = \frac{x}{600} + \frac{1}{2} \] The demand equation is: \[ p = \frac{4,500}{x} \] Setting them equal to each other: \[ \frac{x}{600} + \frac{1}{2} = \frac{4,500}{x} \] To solve for \( x \), first eliminate the fractions by multiplying through by \( 600x \): \[ x^2 + 300x = 2 \cdot 4,500 \] \[ x^2 + 300x - 9000 = 0 \] Next, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1 \), \( b = 300 \), and \( c = -9000 \): \[ x = \frac{-300 \pm \sqrt{300^2 - 4 \cdot 1 \cdot (-9000)}}{2 \cdot 1} \] \[ x = \frac{-300 \pm \sqrt{90000 + 36000}}{2} \] \[ x = \frac{-300 \pm \sqrt{126000}}{2} \] Calculating further: \[ \sqrt{126000} \approx 355.55 \] \[ x = \frac{-300 \pm 355.55}{2} \] Taking the positive root: \[ x \approx \frac{55.55}{2} \approx 27.78 \] We round to the nearest whole number, giving us \( x = 28 \) units. Now, substitute \( x \) back into either equation to find the price: Using the demand equation: \[ p = \frac{4500}{28} \approx 160.71 \] Thus, the units required for supply to equal demand is \( 28 \) bottles, and the price per bottle at that point is approximately \( \$160.71 \).
