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Sagot :
Certainly! Let's go through the problem step-by-step.
### Part A: Find the Marginal Profit Function
The given revenue and cost functions are as follows:
[tex]\[ R(q) = -q^3 + 370q^2 \][/tex]
[tex]\[ C(q) = 270 + 18q \][/tex]
The profit function [tex]\( P(q) \)[/tex] can be defined as the difference between the revenue and the cost functions:
[tex]\[ P(q) = R(q) - C(q) \][/tex]
[tex]\[ P(q) = (-q^3 + 370q^2) - (270 + 18q) \][/tex]
[tex]\[ P(q) = -q^3 + 370q^2 - 270 - 18q \][/tex]
To find the marginal profit function [tex]\( MP(q) \)[/tex], we need to differentiate the profit function [tex]\( P(q) \)[/tex] with respect to [tex]\( q \)[/tex]:
[tex]\[ MP(q) = \frac{d}{dq}[-q^3 + 370q^2 - 270 - 18q] \][/tex]
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
Thus, the simplified expression for the marginal profit function is:
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
### Part B: Number of Items to Maximize Profits
To maximize the profit, we need to find the critical points of the marginal profit function by setting [tex]\( MP(q) \)[/tex] to zero and solving for [tex]\( q \)[/tex]:
[tex]\[ -3q^2 + 740q - 18 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula [tex]\( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -3 \)[/tex], [tex]\( b = 740 \)[/tex], and [tex]\( c = -18 \)[/tex]:
[tex]\[ q = \frac{-740 \pm \sqrt{740^2 - 4(-3)(-18)}}{2(-3)} \][/tex]
Calculating the discriminant:
[tex]\[ 740^2 - 4 \cdot (-3) \cdot (-18) = 547600 - 216 = 547384 \][/tex]
So,
[tex]\[ q = \frac{-740 \pm \sqrt{547384}}{-6} \][/tex]
The solutions for [tex]\( q \)[/tex] are:
[tex]\[ q = \frac{740 \pm \sqrt{547384}}{6} \][/tex]
Simplifying further:
[tex]\[ q = \frac{740 \pm 739.88}{6} \][/tex]
This gives us two solutions:
[tex]\[ q_1 = \frac{740 + 739.88}{6} = \frac{1479.88}{6} \approx 246.64 \][/tex]
[tex]\[ q_2 = \frac{740 - 739.88}{6} = \frac{0.12}{6} \approx 0.02 \][/tex]
Thus, to maximize profits, the company must sell approximately [tex]\( 246.64 \)[/tex] hundred units, which means [tex]\( 246.64 \times 100 = 24664 \)[/tex] units.
### Conclusion
A) The simplified expression for the marginal profit function is:
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
B) To maximize profits, the company needs to sell approximately [tex]\( 246.64 \)[/tex] hundred units, or [tex]\( 24664 \)[/tex] units.
### Part A: Find the Marginal Profit Function
The given revenue and cost functions are as follows:
[tex]\[ R(q) = -q^3 + 370q^2 \][/tex]
[tex]\[ C(q) = 270 + 18q \][/tex]
The profit function [tex]\( P(q) \)[/tex] can be defined as the difference between the revenue and the cost functions:
[tex]\[ P(q) = R(q) - C(q) \][/tex]
[tex]\[ P(q) = (-q^3 + 370q^2) - (270 + 18q) \][/tex]
[tex]\[ P(q) = -q^3 + 370q^2 - 270 - 18q \][/tex]
To find the marginal profit function [tex]\( MP(q) \)[/tex], we need to differentiate the profit function [tex]\( P(q) \)[/tex] with respect to [tex]\( q \)[/tex]:
[tex]\[ MP(q) = \frac{d}{dq}[-q^3 + 370q^2 - 270 - 18q] \][/tex]
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
Thus, the simplified expression for the marginal profit function is:
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
### Part B: Number of Items to Maximize Profits
To maximize the profit, we need to find the critical points of the marginal profit function by setting [tex]\( MP(q) \)[/tex] to zero and solving for [tex]\( q \)[/tex]:
[tex]\[ -3q^2 + 740q - 18 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula [tex]\( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -3 \)[/tex], [tex]\( b = 740 \)[/tex], and [tex]\( c = -18 \)[/tex]:
[tex]\[ q = \frac{-740 \pm \sqrt{740^2 - 4(-3)(-18)}}{2(-3)} \][/tex]
Calculating the discriminant:
[tex]\[ 740^2 - 4 \cdot (-3) \cdot (-18) = 547600 - 216 = 547384 \][/tex]
So,
[tex]\[ q = \frac{-740 \pm \sqrt{547384}}{-6} \][/tex]
The solutions for [tex]\( q \)[/tex] are:
[tex]\[ q = \frac{740 \pm \sqrt{547384}}{6} \][/tex]
Simplifying further:
[tex]\[ q = \frac{740 \pm 739.88}{6} \][/tex]
This gives us two solutions:
[tex]\[ q_1 = \frac{740 + 739.88}{6} = \frac{1479.88}{6} \approx 246.64 \][/tex]
[tex]\[ q_2 = \frac{740 - 739.88}{6} = \frac{0.12}{6} \approx 0.02 \][/tex]
Thus, to maximize profits, the company must sell approximately [tex]\( 246.64 \)[/tex] hundred units, which means [tex]\( 246.64 \times 100 = 24664 \)[/tex] units.
### Conclusion
A) The simplified expression for the marginal profit function is:
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
B) To maximize profits, the company needs to sell approximately [tex]\( 246.64 \)[/tex] hundred units, or [tex]\( 24664 \)[/tex] units.
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