Write an equation using $R(t)=50t e^{-0.1t}+200$ and your target revenue from Hint 1. Then subtract 200 from both sides and divide by 50 to simplify before deciding how to solve.

After simplifying, notice that $t$ appears both outside and inside the exponential. Instead of trying to solve exactly, plug in each answer choice for $t$ (or graph the function) and compare the resulting revenue to your target.

When you compare the values of $R(t)$ for the choices, focus on which ones are below or above the target. Because the revenue is increasing in this time range, which smallest $t$ makes $R(t)$ approximately equal to your target revenue?

In Desmos, enter the revenue model as y1 = 50x*e^(-0.1x) + 200. Then enter the constant target revenue as y2 = 350 to represent \$350,000 dollars.

Adjust the viewing window so you can see $x$ from about 0 to 10 and $y$ from about 200 to 400. Look for where the curve $y_1$ first crosses the horizontal line $y_2$ as $x$ increases from 0, and tap that intersection to see its $x$-coordinate.

Note the $x$-value of that first intersection (in months) and choose the answer option whose value of $t$ is closest to that $x$-coordinate.

The function $R(t)$ gives revenue in thousands of dollars, so $R(t)=350$ means a revenue of $350,000 dollars.

Break-even happens when the revenue first reaches $350,000, so we need to solve

Using the given model,

Start with

Subtract 200 from both sides:

Divide both sides by 50:

Here $t$ appears both outside and inside the exponential, so there is no simple algebraic way (on the SAT) to isolate $t$ exactly.

Because we cannot easily solve $t e^{-0.1t} = 3$ exactly, use the answer choices as guesses and plug them into the original function

with your calculator.

We are looking for:

• A value of $t$ for which $R(t)$ is about $350$ (thousand dollars), and
• The smallest such $t$, because the question asks when the revenue first reaches $350,000.

Note that for $0 < t < 10$, the revenue is increasing, and all the answer choices are less than 10, so once $R(t)$ passes 350, it will stay above 350 in this range.

Now test the answer choices in $R(t) = 50t e^{-0.1t} + 200$ (use your calculator for the exponentials):

• For $t = 3.2$:

Using $e^{-0.32} \approx 0.73$ gives

or about \$316,000, which is **below** \\$350,000.

• For $t = 6.5$:

Using $e^{-0.65} \approx 0.52$ gives

or about \$371,000, which is **above** \\$350,000.

So the break-even time is between $3.2$ and $6.5$ months. Check the remaining options in this range:

• For $t = 4.9$:

Using $e^{-0.49} \approx 0.61$ gives

or about \$350,000.

• For $t = 9.8$:

Using $e^{-0.98} \approx 0.38$ gives

or about \$384,000, which is well **above** \\$350,000 and much later.

The first time the revenue reaches about \$350,000 is at$t \approx 4.9$months, so the correct answer is$4.9$.