To determine the selling price that maximizes the profit, follow these steps:
1. Define the Revenue Function:
The revenue [tex]\( R(x) \)[/tex] comes from selling [tex]\( x \)[/tex] items at the price [tex]\( p(x) \)[/tex], so:
[tex]\[
R(x) = x \cdot p(x) = x \cdot (61 - 0.07x) = 61x - 0.07x^2
\][/tex]
2. Define the Profit Function:
The profit [tex]\( P(x) \)[/tex] is the revenue [tex]\( R(x) \)[/tex] minus the cost [tex]\( C(x) \)[/tex], so:
[tex]\[
P(x) = R(x) - C(x) = (61x - 0.07x^2) - (1325 + 25x - 0.02x^2)
\][/tex]
Simplify the profit function:
[tex]\[
P(x) = 61x - 0.07x^2 - 1325 - 25x + 0.02x^2
\][/tex]
[tex]\[
P(x) = (61x - 25x) + (-0.07x^2 + 0.02x^2) - 1325
\][/tex]
[tex]\[
P(x) = 36x - 0.05x^2 - 1325
\][/tex]
3. Find the Critical Points:
To maximize profit, find the critical points by taking the derivative of [tex]\( P(x) \)[/tex] and setting it to zero:
[tex]\[
P'(x) = \frac{d}{dx}(36x - 0.05x^2 - 1325)
\][/tex]
[tex]\[
P'(x) = 36 - 0.1x
\][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[
36 - 0.1x = 0
\][/tex]
[tex]\[
0.1x = 36
\][/tex]
[tex]\[
x = 360
\][/tex]
4. Confirm that it is a Maximum:
Since this is a quadratic function and the coefficient of [tex]\( x^2 \)[/tex] is negative, the critical point corresponds to a maximum.
5. Find the Selling Price at [tex]\( x = 360 \)[/tex]:
Substitute [tex]\( x = 360 \)[/tex] into the price function [tex]\( p(x) \)[/tex] to determine the selling price:
[tex]\[
p(360) = 61 - 0.07 \times 360
\][/tex]
[tex]\[
p(360) = 61 - 25.2
\][/tex]
[tex]\[
p(360) = 35.8
\][/tex]
The selling price that maximizes the profit is:
[tex]\[
\boxed{35.80}
\][/tex]
