You operate a gaming Web site, where users must pay a small fee to log on. When you charged $4 the demand was 510 log-ons per month. When you lowered the price to $3.50, the demand increased to 765 log-ons per month.
(a) Construct a linear demand function for your web site and hence obtain the monthly revenue R as a function of the log-on fee x.
q(x) =
1
Your answer is correct.
R(x) =
2
Your answer is correct.
(b) Your Internet provider charges you a monthly fee of $30 to maintain your site. Express your monthly profit P as a function of the log-on fee x.
P(x) =
3
Your answer is correct.
Determine the log-on fee you should charge to obtain the largest possible monthly profit.
$ 4Your answer is incorrect.
What is the largest possible monthly profit?
To construct a linear demand function for your web site, we can use the price-demand equation. The price-demand equation relates the price \(x\) and the demand \(q(x)\).
Given that the demand when the price is $4 is 510 log-ons per month, and the demand when the price is $3.50 is 765 log-ons per month, we can use these two data points to find the slope (\(m\)) and y-intercept (\(b\)) of the linear demand function.
Using the slope-intercept form of a linear equation, which is \(y = mx + b\), where \(y\) represents the demand \(q(x)\) and \(x\) represents the price \(x\), we can rearrange the equation to find the demand equation:
\[q(x) = mx + b\]
Now, let's find the slope (\(m\)):
\[m = \frac{{\text{{change in }} q}}{{\text{{change in }} x}} = \frac{{765 - 510}}{{3.5 - 4}}\]
Simplifying the calculation:
\[m = \frac{{255}}{{-0.50}} = -510\]
Next, let's find the y-intercept (\(b\)):
\[b = q(x) - mx = 510 - (-510) \times 4 = 510 + 2,040 = 2,550\]
Thus, the demand function for your web site is:
\[q(x) = -510x + 2,550\]
Now, let's find the monthly revenue (\(R(x)\)) as a function of the log-on fee \(x\). The revenue is calculated by multiplying the demand (\(q(x)\)) by the log-on fee (\(x\)):
\[R(x) = x \times q(x) = x \times (-510x + 2,550)\]
To find the monthly profit (\(P(x)\)), we need to subtract the monthly cost of $30 from the monthly revenue:
\[P(x) = R(x) - 30 = -510x^2 + 2,550x - 30\]
To determine the log-on fee that will maximize the monthly profit, we need to find the x-value (price) at the vertex of the quadratic equation \(P(x)\). The optimal price occurs at the vertex of the parabola. We can use the formula for the x-value of the vertex, which is \(-\frac{b}{2a}\), where \(a\) is the coefficient of \(x^2\) (\(-510\)) and \(b\) is the coefficient of \(x\) (\(2,550\)):
\[x = -\frac{2550}{2(-510)} = -\frac{2550}{-1020} = 2.5\]
Therefore, to obtain the largest possible monthly profit, you should charge a log-on fee of $2.50.
To find the largest possible monthly profit, substitute the optimal price (\(x = 2.5\)) into the profit equation:
\[P(x) = -510x^2 + 2,550x - 30\]
\[P(2.5) = -510(2.5)^2 + 2,550(2.5) - 30\]
\[P(2.5) = -510(6.25) + 6,375 - 30\]
\[P(2.5) = -3,187.50 + 6,375 - 30\]
\[P(2.5) = 3,157.50\]
Therefore, the largest possible monthly profit is $3,157.50.