To determine the number of speakers that should be sold to maximize profit given the profit function [tex]\( P(x) = 3 + x - 5x^2 \)[/tex], we need to follow a series of steps involving calculus.
1. Identify the Profit Function:
The profit function is given by:
[tex]\[
P(x) = 3 + x - 5x^2
\][/tex]
where [tex]\( x \)[/tex] represents the number of thousand speakers sold.
2. Find the First Derivative:
To find the critical points, we first need to compute the first derivative of [tex]\( P(x) \)[/tex]. The first derivative, [tex]\( P'(x) \)[/tex], represents the rate of change of profit with respect to the number of speakers sold.
[tex]\[
P'(x) = \frac{d}{dx} (3 + x - 5x^2) = 1 - 10x
\][/tex]
3. Set the First Derivative to Zero:
The critical points occur where the first derivative is zero. We solve for [tex]\( x \)[/tex] in the equation:
[tex]\[
1 - 10x = 0
\][/tex]
[tex]\[
10x = 1
\][/tex]
[tex]\[
x = \frac{1}{10}
\][/tex]
4. Find the Second Derivative:
To determine whether this critical point is a maximum or minimum, we can use the second derivative test. We compute the second derivative of [tex]\( P(x) \)[/tex]:
[tex]\[
P''(x) = \frac{d}{dx} (1 - 10x) = -10
\][/tex]
5. Evaluate the Second Derivative at the Critical Point:
We evaluate the second derivative at [tex]\( x = \frac{1}{10} \)[/tex]:
[tex]\[
P''\left( \frac{1}{10} \right) = -10
\][/tex]
Since [tex]\( P''\left( \frac{1}{10} \right) = -10 \)[/tex] is negative, the critical point [tex]\( x = \frac{1}{10} \)[/tex] is a local maximum.
6. Conclude the Optimal Number of Speakers:
Therefore, to maximize the profit, the number of speakers [tex]\( x \)[/tex] that should be sold is [tex]\( \frac{1}{10} \)[/tex] thousand, which is equivalent to 100 speakers.
Therefore, to maximize profit, you should sell [tex]\( \boxed{100} \)[/tex] speakers.
