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To determine the range of laptops that the company should produce per day to make a profit, we first need to formulate the profit function. The profit function [tex]\( P(x) \)[/tex] is given by the difference between the revenue function [tex]\( R(x) \)[/tex] and the cost function [tex]\( C(x) \)[/tex]:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
Given:
- The cost function is [tex]\( C(x) = 3x^2 - 10x + 200 \)[/tex]
- The revenue function is [tex]\( R(x) = -2x^2 + 100x + 50 \)[/tex]
The profit function thus becomes:
[tex]\[ P(x) = (-2x^2 + 100x + 50) - (3x^2 - 10x + 200) \][/tex]
Simplify the expression:
[tex]\[ P(x) = -2x^2 + 100x + 50 - 3x^2 + 10x - 200 \][/tex]
[tex]\[ P(x) = -2x^2 - 3x^2 + 100x + 10x + 50 - 200 \][/tex]
[tex]\[ P(x) = -5x^2 + 110x - 150 \][/tex]
To determine when the company makes a profit, we need to find the values of [tex]\( x \)[/tex] (the number of laptops produced per day) for which [tex]\( P(x) > 0 \)[/tex]. In other words:
[tex]\[ -5x^2 + 110x - 150 > 0 \][/tex]
This is a quadratic inequality. To solve it, we first find the roots of the corresponding quadratic equation by setting the profit function to zero:
[tex]\[ -5x^2 + 110x - 150 = 0 \][/tex]
We can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = -5 \)[/tex], [tex]\( b = 110 \)[/tex], and [tex]\( c = -150 \)[/tex]:
[tex]\[ x = \frac{-110 \pm \sqrt{110^2 - 4(-5)(-150)}}{2(-5)} \][/tex]
[tex]\[ x = \frac{-110 \pm \sqrt{12100 - 3000}}{-10} \][/tex]
[tex]\[ x = \frac{-110 \pm \sqrt{9100}}{-10} \][/tex]
[tex]\[ x = \frac{-110 \pm 95.4}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{-110 + 95.4}{-10} = 1.46 \][/tex]
[tex]\[ x_2 = \frac{-110 - 95.4}{-10} = 20.54 \][/tex]
Since the company can only produce whole laptops, we round these values to the nearest whole number. Therefore, the possible range of [tex]\( x \)[/tex] is from 2 to 20.
To be certain that this range generates profit, we examine our inequality around these critical points:
- For [tex]\( x \in (2, 20) \)[/tex], the profit function [tex]\( P(x) > 0 \)[/tex].
- Outside this interval, the company does not generate profit.
Thus, the company should produce between 2 and 20 laptops per day to make a profit. This means the range of production for generating profit is:
[tex]\[ \boxed{[2, 20]} \][/tex]
[tex]\[ P(x) = R(x) - C(x) \][/tex]
Given:
- The cost function is [tex]\( C(x) = 3x^2 - 10x + 200 \)[/tex]
- The revenue function is [tex]\( R(x) = -2x^2 + 100x + 50 \)[/tex]
The profit function thus becomes:
[tex]\[ P(x) = (-2x^2 + 100x + 50) - (3x^2 - 10x + 200) \][/tex]
Simplify the expression:
[tex]\[ P(x) = -2x^2 + 100x + 50 - 3x^2 + 10x - 200 \][/tex]
[tex]\[ P(x) = -2x^2 - 3x^2 + 100x + 10x + 50 - 200 \][/tex]
[tex]\[ P(x) = -5x^2 + 110x - 150 \][/tex]
To determine when the company makes a profit, we need to find the values of [tex]\( x \)[/tex] (the number of laptops produced per day) for which [tex]\( P(x) > 0 \)[/tex]. In other words:
[tex]\[ -5x^2 + 110x - 150 > 0 \][/tex]
This is a quadratic inequality. To solve it, we first find the roots of the corresponding quadratic equation by setting the profit function to zero:
[tex]\[ -5x^2 + 110x - 150 = 0 \][/tex]
We can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = -5 \)[/tex], [tex]\( b = 110 \)[/tex], and [tex]\( c = -150 \)[/tex]:
[tex]\[ x = \frac{-110 \pm \sqrt{110^2 - 4(-5)(-150)}}{2(-5)} \][/tex]
[tex]\[ x = \frac{-110 \pm \sqrt{12100 - 3000}}{-10} \][/tex]
[tex]\[ x = \frac{-110 \pm \sqrt{9100}}{-10} \][/tex]
[tex]\[ x = \frac{-110 \pm 95.4}{-10} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{-110 + 95.4}{-10} = 1.46 \][/tex]
[tex]\[ x_2 = \frac{-110 - 95.4}{-10} = 20.54 \][/tex]
Since the company can only produce whole laptops, we round these values to the nearest whole number. Therefore, the possible range of [tex]\( x \)[/tex] is from 2 to 20.
To be certain that this range generates profit, we examine our inequality around these critical points:
- For [tex]\( x \in (2, 20) \)[/tex], the profit function [tex]\( P(x) > 0 \)[/tex].
- Outside this interval, the company does not generate profit.
Thus, the company should produce between 2 and 20 laptops per day to make a profit. This means the range of production for generating profit is:
[tex]\[ \boxed{[2, 20]} \][/tex]
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