Question

the profitability, p, of a popular restaurant franchise can be modeled by the function $p(t) = t^3 - 9t^2 + 24t^2 - 20t$, where t is the number of months since the restaurant opened. how many months after the franchise opens will it begin to show a profit?
5 months
3 months
2 months
1 month

Explanation:

Step1: Understand the profit function

The profit function is \( P(t) = t^3 - 9t^2 + 24t - 20 \). We need to find the smallest \( t>0 \) (in months) where \( P(t)>0 \), meaning the restaurant starts to show a profit.

Step2: Test each option by substituting \( t \) into the profit function

- **For \( t = 1 \) month:**
Substitute \( t = 1 \) into \( P(t) \):
\( P(1)=(1)^3 - 9(1)^2 + 24(1) - 20 \)
\( = 1 - 9 + 24 - 20 \)
\( = (1 + 24) - (9 + 20) \)
\( = 25 - 29 \)
\( = -4 \)
Since \( P(1)= - 4<0 \), the restaurant is not profitable at \( t = 1 \) month.

- **For \( t = 2 \) months:**
Substitute \( t = 2 \) into \( P(t) \):
\( P(2)=(2)^3 - 9(2)^2 + 24(2) - 20 \)
\( = 8 - 36 + 48 - 20 \)
\( = (8 + 48) - (36 + 20) \)
\( = 56 - 56 \)
\( = 0 \)
At \( t = 2 \) months, the profit is zero (break - even point), not yet showing a profit.

- **For \( t = 3 \) months:**
Substitute \( t = 3 \) into \( P(t) \):
\( P(3)=(3)^3 - 9(3)^2 + 24(3) - 20 \)
\( = 27 - 81 + 72 - 20 \)
\( = (27 + 72) - (81 + 20) \)
\( = 99 - 101 \)
\( = -2 \)
Since \( P(3)= - 2<0 \), the restaurant is not profitable at \( t = 3 \) months.

- **For \( t = 5 \) months:**
Substitute \( t = 5 \) into \( P(t) \):
\( P(5)=(5)^3 - 9(5)^2 + 24(5) - 20 \)
\( = 125 - 225 + 120 - 20 \)
\( = (125 + 120) - (225 + 20) \)
\( = 245 - 245 \)
Wait, that's incorrect. Let's recalculate:
\( P(5)=125- 9\times25 + 24\times5-20=125 - 225+120 - 20=(125 + 120)-(225 + 20)=245 - 245 = 0 \)? No, wait, maybe we made a mistake. Wait, let's re - evaluate the function. Wait, maybe we should factor the function or try another value. Wait, perhaps we made an error in calculation. Let's try \( t = 4 \) (even though it's not an option, but to check the trend). \( P(4)=64-9\times16 + 24\times4-20=64 - 144 + 96 - 20=(64 + 96)-(144 + 20)=160 - 164=-4 \). Wait, maybe the options are wrong? Wait, no, let's re - check the original function. The function is \( P(t)=t^{3}-9t^{2}+24t - 20 \). Let's try to factor it. We can use the rational root theorem. The possible rational roots are factors of 20 over factors of 1, i.e., \( \pm1,\pm2,\pm4,\pm5,\pm10,\pm20 \). We saw that \( P(2) = 8-36 + 48 - 20=0 \), \( P(1)=1 - 9+24 - 20=-4 \), \( P(5)=125-225 + 120 - 20=0 \), \( P(4)=64-144 + 96 - 20=-4 \), \( P(3)=27-81 + 72 - 20=-2 \). Wait, this suggests that the function has roots at \( t = 2 \), \( t = 5 \) and maybe another root? Wait, let's perform polynomial division. Divide \( t^{3}-9t^{2}+24t - 20 \) by \( (t - 2) \). Using synthetic division:
2 | 1 - 9 24 - 20
2 - 14 20
1 - 7 10 0
So the function factors as \( (t - 2)(t^{2}-7t + 10)=(t - 2)(t - 2)(t - 5)=(t - 2)^2(t - 5) \). Ah! So \( P(t)=(t - 2)^2(t - 5) \). Now, we analyze the sign of \( P(t) \):
- For \( t<2 \): Let's take \( t = 1 \), \( P(1)=(1 - 2)^2(1 - 5)=(-1)^2\times(-4)=-4<0 \)
- For \( 2<t<5 \): Let's take \( t = 3 \), \( P(3)=(3 - 2)^2(3 - 5)=(1)^2\times(-2)=-2<0 \)
- For \( t>5 \): Let's take \( t = 6 \), \( P(6)=(6 - 2)^2(6 - 5)=16\times1 = 16>0 \)

Wait, but the options given are 5, 3, 2, 1. According to our factored form \( P(t)=(t - 2)^2(t - 5) \), \( P(t)>0 \) when \( t>5 \) (since \( (t - 2)^2\geq0 \) for all real \( t \), and \( (t - 5)>0 \) when \( t>5 \)). But the options do not have \( t>5 \). There must be a mistake in the problem statement or the options. But among the given options, when \( t = 5 \), \( P(5)=(5 - 2)^2(5 - 5)=0 \), when \( t>5 \), it's positive. But since the options are as given, maybe there is a miscalculation. Wait, maybe the original function was \( P(t)=t^{3}-9t^{2}+24t + 20 \)? No, the user wrote \( P(t)=t^{3}-9t^{2}+24t - 20 \). Alternatively, maybe we misread the function. If the function is \( P(t)=-t^{3}+9t^{2}-24t + 20 \) (profit as a negative of the given function), let's check. Then \( P(1)=-1 + 9-24 + 20 = 4>0 \)? No, that doesn't make sense. Wait, the original problem says "the profitability, \( P \), of a popular restaurant franchise can be modeled by the function \( P(t)=t^{3}-9t^{2}+24t - 20 \), where \( t \) is the number of months since the restaurant opened. How many months after the franchise opens will it begin to show a profit?"

From our factoring, \( P(t)=(t - 2)^2(t - 5) \). So \( P(t)=0 \) at \( t = 2 \) and \( t = 5 \), and \( P(t)<0 \) for \( t\in(0,2)\cup(2,5) \), \( P(t)>0 \) for \( t>5 \). So among the given options, none of them is correct in the strict sense. But maybe there was a typo in the function. If the function was \( P(t)=-t^{3}+9t^{2}-24t + 20 \), let's check:

\( P(1)=-1 + 9-24 + 20 = 4>0 \), but that contradicts. Alternatively, if the function was \( P(t)=t^{3}-9t^{2}+24t + 20 \), \( P(1)=1 - 9+24 + 20 = 36>0 \). But according to the original problem, the options are 5,3,2,1.

Wait, maybe we made a mistake in factoring. Let's re - do the synthetic division:

For \( t = 2 \):

Coefficients: 1 (t³), - 9 (t²), 24 (t), - 20 (constant)

Bring down the 1.

Multiply 1 by 2: 2. Add to - 9: - 7.

Multiply - 7 by 2: - 14. Add to 24: 10.

Multiply 10 by 2: 20. Add to - 20: 0. So the quotient is \( t^{2}-7t + 10 \), which factors into \( (t - 2)(t - 5) \). So \( P(t)=(t - 2)^2(t - 5) \). So the profit is positive when \( t>5 \). Since 5 is an option, and at \( t = 5 \), \( P(5)=0 \), and for \( t>5 \), \( P(t)>0 \). So the restaurant begins to show a profit after 5 months (when \( t>5 \), and at \( t = 5 \), it's break - even). So among the given options, the answer should be 5 months.

Answer:

5 months (Note: There is a slight ambiguity as at \( t = 5 \) the profit is zero, and it becomes positive for \( t>5 \), but among the given options, 5 months is the closest as it is the point where the profit transitions from non - positive to positive as \( t \) increases beyond 5.)