Answer :
To determine which selling prices of lawnmowers result in a profit that is greater than or equal to zero, we need to evaluate the profit, defined as the difference between the revenue and cost functions, [tex]\( \pi(x) = R(x) - C(x) \)[/tex], for each given price.
The given revenue and cost functions are:
[tex]\[ R(x) = -2.075x^2 + 768x \][/tex]
[tex]\[ C(x) = -134.625x + 84037.5 \][/tex]
We'll first evaluate the profit at each given price: 120, 135, 285, 300, and 335.
### Evaluation of Profit at Given Prices
1. Price [tex]\( x = 120 \)[/tex]
[tex]\[ R(120) = -2.075(120)^2 + 768(120) \][/tex]
[tex]\[ C(120) = -134.625(120) + 84037.5 \][/tex]
[tex]\[ \pi(120) = R(120) - C(120) = -5602.5 \][/tex]
The profit is [tex]\(-5602.5\)[/tex], so this price does not yield a non-negative profit.
2. Price [tex]\( x = 135 \)[/tex]
[tex]\[ R(135) = -2.075(135)^2 + 768(135) \][/tex]
[tex]\[ C(135) = -134.625(135) + 84037.5 \][/tex]
[tex]\[ \pi(135) = R(135) - C(135) = 0.0 \][/tex]
The profit is [tex]\(0.0\)[/tex], which is a non-negative profit.
3. Price [tex]\( x = 285 \)[/tex]
[tex]\[ R(285) = -2.075(285)^2 + 768(285) \][/tex]
[tex]\[ C(285) = -134.625(285) + 84037.5 \][/tex]
[tex]\[ \pi(285) = R(285) - C(285) = 4668.75 \][/tex]
The profit is [tex]\(4668.75\)[/tex], which is a non-negative profit.
4. Price [tex]\( x = 300 \)[/tex]
[tex]\[ R(300) = -2.075(300)^2 + 768(300) \][/tex]
[tex]\[ C(300) = -134.625(300) + 84037.5 \][/tex]
[tex]\[ \pi(300) = R(300) - C(300) = -2.9103830456733704 \times 10^{-11} \][/tex]
The profit is [tex]\(-2.9103830456733704 \times 10^{-11}\)[/tex], which is effectively zero but slightly negative due to numerical precision.
5. Price [tex]\( x = 335 \)[/tex]
[tex]\[ R(335) = -2.075(335)^2 + 768(335) \][/tex]
[tex]\[ C(335) = -134.625(335) + 84037.5 \][/tex]
[tex]\[ \pi(335) = R(335) - C(335) = -14525.00000000003 \][/tex]
The profit is [tex]\(-14525.00000000003\)[/tex], which is negative.
### Conclusion
The prices that result in a profit that is greater than or equal to zero are:
135, 285.
So, the correct answers are:
- 135
- 285
The given revenue and cost functions are:
[tex]\[ R(x) = -2.075x^2 + 768x \][/tex]
[tex]\[ C(x) = -134.625x + 84037.5 \][/tex]
We'll first evaluate the profit at each given price: 120, 135, 285, 300, and 335.
### Evaluation of Profit at Given Prices
1. Price [tex]\( x = 120 \)[/tex]
[tex]\[ R(120) = -2.075(120)^2 + 768(120) \][/tex]
[tex]\[ C(120) = -134.625(120) + 84037.5 \][/tex]
[tex]\[ \pi(120) = R(120) - C(120) = -5602.5 \][/tex]
The profit is [tex]\(-5602.5\)[/tex], so this price does not yield a non-negative profit.
2. Price [tex]\( x = 135 \)[/tex]
[tex]\[ R(135) = -2.075(135)^2 + 768(135) \][/tex]
[tex]\[ C(135) = -134.625(135) + 84037.5 \][/tex]
[tex]\[ \pi(135) = R(135) - C(135) = 0.0 \][/tex]
The profit is [tex]\(0.0\)[/tex], which is a non-negative profit.
3. Price [tex]\( x = 285 \)[/tex]
[tex]\[ R(285) = -2.075(285)^2 + 768(285) \][/tex]
[tex]\[ C(285) = -134.625(285) + 84037.5 \][/tex]
[tex]\[ \pi(285) = R(285) - C(285) = 4668.75 \][/tex]
The profit is [tex]\(4668.75\)[/tex], which is a non-negative profit.
4. Price [tex]\( x = 300 \)[/tex]
[tex]\[ R(300) = -2.075(300)^2 + 768(300) \][/tex]
[tex]\[ C(300) = -134.625(300) + 84037.5 \][/tex]
[tex]\[ \pi(300) = R(300) - C(300) = -2.9103830456733704 \times 10^{-11} \][/tex]
The profit is [tex]\(-2.9103830456733704 \times 10^{-11}\)[/tex], which is effectively zero but slightly negative due to numerical precision.
5. Price [tex]\( x = 335 \)[/tex]
[tex]\[ R(335) = -2.075(335)^2 + 768(335) \][/tex]
[tex]\[ C(335) = -134.625(335) + 84037.5 \][/tex]
[tex]\[ \pi(335) = R(335) - C(335) = -14525.00000000003 \][/tex]
The profit is [tex]\(-14525.00000000003\)[/tex], which is negative.
### Conclusion
The prices that result in a profit that is greater than or equal to zero are:
135, 285.
So, the correct answers are:
- 135
- 285