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Sagot :
Absolutely, let's tackle the given problem with careful consideration of each part. We'll go through parts (a) through (c) step-by-step.
### Part (a): Sketching the Total Revenue (TR) Graph
To sketch the TR against [tex]\( Q \)[/tex] on the interval [tex]\( 0 \leq Q \leq 10 \)[/tex], let's plot the given functions:
1. For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ TR = 0.2 Q^3 \][/tex]
2. For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ TR = -4 Q^2 + 55 Q - 150 \][/tex]
Let's identify key points:
- At [tex]\( Q = 0 \)[/tex]:
- [tex]\( TR = 0.2 \times 0^3 = 0 \)[/tex]
- At [tex]\( Q = 5 \)[/tex]:
- Using [tex]\( TR = 0.2 Q^3 \)[/tex]:
[tex]\[ TR = 0.2 \times 5^3 = 0.2 \times 125 = 25 \][/tex]
- Using [tex]\( TR = -4Q^2 + 55Q - 150 \)[/tex]:
[tex]\[ TR = -4 \times 5^2 + 55 \times 5 - 150 = -100 + 275 - 150 = 25 \][/tex]
(Both functions yield the same value at [tex]\( Q = 5 \)[/tex], ensuring continuity.)
- At [tex]\( Q = 10 \)[/tex]:
- [tex]\( TR = -4 \times 10^2 + 55 \times 10 - 150 = -400 + 550 - 150 = 0 \)[/tex]
A rough sketch of these segments will show [tex]\( TR \)[/tex] increasing as [tex]\( Q \)[/tex] increases from 0 to 5, reaching a peak, and then decreasing as [tex]\( Q \)[/tex] increases from 5 to 10.
### Part (b): Finding the Maximum Revenue and Corresponding [tex]\( Q \)[/tex]
For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ TR = 0.2 Q^3 \][/tex]
To find the maximum, take the derivative and set it to zero:
[tex]\[ \frac{d(TR)}{dQ} = 0.6 Q^2 \][/tex]
Setting the derivative to zero:
[tex]\[ 0.6 Q^2 = 0 \implies Q = 0 \][/tex]
_Check boundary [tex]\( Q=5 \)[/tex]_:
[tex]\[ TR(5) = 0.2 \times 5^3 = 25 \][/tex]
For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ TR = -4 Q^2 + 55 Q - 150 \][/tex]
Take the derivative:
[tex]\[ \frac{d(TR)}{dQ} = -8Q + 55 \][/tex]
Set the derivative to zero and solve for [tex]\( Q \)[/tex]:
[tex]\[ -8Q + 55 = 0 \][/tex]
[tex]\[ 8Q = 55 \][/tex]
[tex]\[ Q = \frac{55}{8} = 6.875 \][/tex]
Evaluate [tex]\( TR \)[/tex] at [tex]\( Q = 6.875 \)[/tex]:
[tex]\[ TR(6.875) = -4 \times (6.875)^2 + 55 \times 6.875 - 150 \][/tex]
[tex]\[ TR(6.875) \approx 36.719 \][/tex]
Compare the [tex]\( TR \)[/tex] values:
- At [tex]\( Q = 0 \)[/tex], [tex]\( TR = 0 \)[/tex]
- At [tex]\( Q = 5 \)[/tex], [tex]\( TR = 25 \)[/tex]
- At [tex]\( Q = 6.875 \)[/tex], [tex]\( TR \approx 36.719 \)[/tex]
- At [tex]\( Q = 10 \)[/tex], [tex]\( TR = 0 \)[/tex]
The maximum revenue is [tex]\( TR \approx 36.719 \)[/tex] at [tex]\( Q = 6.875 \)[/tex].
### Part (c): Maximizing Marginal Revenue (MR)
Marginal Revenue (MR) is the derivative of TR.
For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ TR = 0.2 Q^3 \][/tex]
[tex]\[ \frac{d(TR)}{dQ} = MR = 0.6 Q^2 \][/tex]
For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ TR = -4 Q^2 + 55 Q - 150 \][/tex]
[tex]\[ \frac{d(TR)}{dQ} = MR = -8Q + 55 \][/tex]
To find where MR is maximum in each segment:
For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ MR = 0.6 Q^2 \][/tex]
As [tex]\( Q \)[/tex] increases from 0 to 5, [tex]\( MR \)[/tex] increases since it's a positive quadratic function.
For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ MR = -8Q + 55 \][/tex]
It's a linear function decreasing with [tex]\( Q \)[/tex]. It reaches maximum at the left boundary:
[tex]\[ Q = 5 \][/tex]
[tex]\[ MR = -8 \times 5 + 55 = 15 \][/tex]
Compare [tex]\( MR \)[/tex] values:
- At [tex]\( Q = 5 \)[/tex] from the second segment: [tex]\( MR = 15 \)[/tex]
- For the first segment, [tex]\( MR \)[/tex] increases with [tex]\( Q \)[/tex], reaching its peak at [tex]\( Q = 5 \)[/tex]:
[tex]\[ MR = 0.6 \times 5^2 = 15 \][/tex]
So, maximum marginal revenue of 15 occurs at [tex]\( Q = 5 \)[/tex].
### Summary:
- (a) We sketched the TR curve based on given functions.
- (b) The maximum revenue [tex]\( TR \approx 36.719 \)[/tex] is achieved at [tex]\( Q = 6.875 \)[/tex].
- (c) The marginal revenue is maximum at [tex]\( Q = 5 \)[/tex] and [tex]\( MR = 15 \)[/tex].
This step-by-step solution covers the graphing, maximum revenue point, and marginal revenue analysis!
### Part (a): Sketching the Total Revenue (TR) Graph
To sketch the TR against [tex]\( Q \)[/tex] on the interval [tex]\( 0 \leq Q \leq 10 \)[/tex], let's plot the given functions:
1. For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ TR = 0.2 Q^3 \][/tex]
2. For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ TR = -4 Q^2 + 55 Q - 150 \][/tex]
Let's identify key points:
- At [tex]\( Q = 0 \)[/tex]:
- [tex]\( TR = 0.2 \times 0^3 = 0 \)[/tex]
- At [tex]\( Q = 5 \)[/tex]:
- Using [tex]\( TR = 0.2 Q^3 \)[/tex]:
[tex]\[ TR = 0.2 \times 5^3 = 0.2 \times 125 = 25 \][/tex]
- Using [tex]\( TR = -4Q^2 + 55Q - 150 \)[/tex]:
[tex]\[ TR = -4 \times 5^2 + 55 \times 5 - 150 = -100 + 275 - 150 = 25 \][/tex]
(Both functions yield the same value at [tex]\( Q = 5 \)[/tex], ensuring continuity.)
- At [tex]\( Q = 10 \)[/tex]:
- [tex]\( TR = -4 \times 10^2 + 55 \times 10 - 150 = -400 + 550 - 150 = 0 \)[/tex]
A rough sketch of these segments will show [tex]\( TR \)[/tex] increasing as [tex]\( Q \)[/tex] increases from 0 to 5, reaching a peak, and then decreasing as [tex]\( Q \)[/tex] increases from 5 to 10.
### Part (b): Finding the Maximum Revenue and Corresponding [tex]\( Q \)[/tex]
For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ TR = 0.2 Q^3 \][/tex]
To find the maximum, take the derivative and set it to zero:
[tex]\[ \frac{d(TR)}{dQ} = 0.6 Q^2 \][/tex]
Setting the derivative to zero:
[tex]\[ 0.6 Q^2 = 0 \implies Q = 0 \][/tex]
_Check boundary [tex]\( Q=5 \)[/tex]_:
[tex]\[ TR(5) = 0.2 \times 5^3 = 25 \][/tex]
For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ TR = -4 Q^2 + 55 Q - 150 \][/tex]
Take the derivative:
[tex]\[ \frac{d(TR)}{dQ} = -8Q + 55 \][/tex]
Set the derivative to zero and solve for [tex]\( Q \)[/tex]:
[tex]\[ -8Q + 55 = 0 \][/tex]
[tex]\[ 8Q = 55 \][/tex]
[tex]\[ Q = \frac{55}{8} = 6.875 \][/tex]
Evaluate [tex]\( TR \)[/tex] at [tex]\( Q = 6.875 \)[/tex]:
[tex]\[ TR(6.875) = -4 \times (6.875)^2 + 55 \times 6.875 - 150 \][/tex]
[tex]\[ TR(6.875) \approx 36.719 \][/tex]
Compare the [tex]\( TR \)[/tex] values:
- At [tex]\( Q = 0 \)[/tex], [tex]\( TR = 0 \)[/tex]
- At [tex]\( Q = 5 \)[/tex], [tex]\( TR = 25 \)[/tex]
- At [tex]\( Q = 6.875 \)[/tex], [tex]\( TR \approx 36.719 \)[/tex]
- At [tex]\( Q = 10 \)[/tex], [tex]\( TR = 0 \)[/tex]
The maximum revenue is [tex]\( TR \approx 36.719 \)[/tex] at [tex]\( Q = 6.875 \)[/tex].
### Part (c): Maximizing Marginal Revenue (MR)
Marginal Revenue (MR) is the derivative of TR.
For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ TR = 0.2 Q^3 \][/tex]
[tex]\[ \frac{d(TR)}{dQ} = MR = 0.6 Q^2 \][/tex]
For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ TR = -4 Q^2 + 55 Q - 150 \][/tex]
[tex]\[ \frac{d(TR)}{dQ} = MR = -8Q + 55 \][/tex]
To find where MR is maximum in each segment:
For [tex]\( 0 \leq Q \leq 5 \)[/tex]:
[tex]\[ MR = 0.6 Q^2 \][/tex]
As [tex]\( Q \)[/tex] increases from 0 to 5, [tex]\( MR \)[/tex] increases since it's a positive quadratic function.
For [tex]\( 5 \leq Q \leq 10 \)[/tex]:
[tex]\[ MR = -8Q + 55 \][/tex]
It's a linear function decreasing with [tex]\( Q \)[/tex]. It reaches maximum at the left boundary:
[tex]\[ Q = 5 \][/tex]
[tex]\[ MR = -8 \times 5 + 55 = 15 \][/tex]
Compare [tex]\( MR \)[/tex] values:
- At [tex]\( Q = 5 \)[/tex] from the second segment: [tex]\( MR = 15 \)[/tex]
- For the first segment, [tex]\( MR \)[/tex] increases with [tex]\( Q \)[/tex], reaching its peak at [tex]\( Q = 5 \)[/tex]:
[tex]\[ MR = 0.6 \times 5^2 = 15 \][/tex]
So, maximum marginal revenue of 15 occurs at [tex]\( Q = 5 \)[/tex].
### Summary:
- (a) We sketched the TR curve based on given functions.
- (b) The maximum revenue [tex]\( TR \approx 36.719 \)[/tex] is achieved at [tex]\( Q = 6.875 \)[/tex].
- (c) The marginal revenue is maximum at [tex]\( Q = 5 \)[/tex] and [tex]\( MR = 15 \)[/tex].
This step-by-step solution covers the graphing, maximum revenue point, and marginal revenue analysis!
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