Question
Ernesto created a mindfulness app called BrainShift. He conducted a survey to determine how the price to download the BrainShift app will affect the number of downloads he sells. The survey indicated that if he charges x dollars per download, he will sell
–
25x+500 downloads per month.
The app store will charge Ernesto 40% of the amount he charges per download. So, Ernesto will earn 60% of the amount he charges per download, or 0.6x dollars, in profit.
Which equation can Ernesto use to calculate the price he can charge per download to earn $1,100 per month in profit?
To the nearest dollar, what is the lowest price Ernesto can charge per download to earn $1,100 per month in profit?
$
Questions
answered
8
Time
elapsed
00 14 31
hr min sec
SmartScore
out of 100
54
Need a break?
–
25x+500 downloads per month.
The app store will charge Ernesto 40% of the amount he charges per download. So, Ernesto will earn 60% of the amount he charges per download, or 0.6x dollars, in profit.
Which equation can Ernesto use to calculate the price he can charge per download to earn $1,100 per month in profit?
To the nearest dollar, what is the lowest price Ernesto can charge per download to earn $1,100 per month in profit?
$
Questions
answered
8
Time
elapsed
00 14 31
hr min sec
SmartScore
out of 100
54
Need a break?
Answers
GPT-4o mini
To find the equation Ernesto can use to calculate the price per download, we'll start with the information given.
1. The number of downloads per month based on the price \(x\) is given by:
\[
D(x) = -25x + 500
\]
2. Ernesto earns 60% of the price per download, which is:
\[
\text{Profit per download} = 0.6x
\]
3. Therefore, total profit per month \(P\) can be calculated as:
\[
P = \text{Profit per download} \times \text{Number of downloads}
\]
Substituting the equations we have:
\[
P = 0.6x \times D(x) = 0.6x \times (-25x + 500)
\]
4. He wants to earn $1,100 per month in profit, so we set up the equation:
\[
0.6x \times (-25x + 500) = 1100
\]
Now let's simplify the equation:
- First, expand the left side:
\[
0.6x(-25x + 500) = -15x^2 + 300x
\]
So the equation becomes:
\[
-15x^2 + 300x = 1100
\]
5. Rearranging this gives us the quadratic equation:
\[
-15x^2 + 300x - 1100 = 0
\]
We can multiply through by -1 to make it standard form:
\[
15x^2 - 300x + 1100 = 0
\]
Now, to find the lowest price Ernesto can charge, we can solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 15\), \(b = -300\), and \(c = 1100\).
6. Calculate \(b^2 - 4ac\):
\[
b^2 - 4ac = (-300)^2 - 4 \cdot 15 \cdot 1100
\]
\[
= 90000 - 66000 = 24000
\]
7. Now apply the quadratic formula:
\[
x = \frac{-(-300) \pm \sqrt{24000}}{2 \cdot 15}
\]
\[
= \frac{300 \pm \sqrt{24000}}{30}
\]
8. Calculate \(\sqrt{24000}\):
\[
\sqrt{24000} \approx 155
\]
(rounding for simplicity)
9. Now substitute back into the formula:
\[
x = \frac{300 \pm 155}{30}
\]
This gives us two potential solutions:
\[
x_1 = \frac{455}{30} \approx 15.17 \quad \text{and} \quad x_2 = \frac{145}{30} \approx 4.83
\]
10. Since we need the lowest price that still meets the requirement of $1100 in profit, we take the higher value (since as x decreases, profit decreases). So we round \(15.17\) to the nearest dollar.
Thus, the lowest price Ernesto can charge per download to earn $1,100 per month in profit is:
\[
\boxed{15}
\]
1. The number of downloads per month based on the price \(x\) is given by:
\[
D(x) = -25x + 500
\]
2. Ernesto earns 60% of the price per download, which is:
\[
\text{Profit per download} = 0.6x
\]
3. Therefore, total profit per month \(P\) can be calculated as:
\[
P = \text{Profit per download} \times \text{Number of downloads}
\]
Substituting the equations we have:
\[
P = 0.6x \times D(x) = 0.6x \times (-25x + 500)
\]
4. He wants to earn $1,100 per month in profit, so we set up the equation:
\[
0.6x \times (-25x + 500) = 1100
\]
Now let's simplify the equation:
- First, expand the left side:
\[
0.6x(-25x + 500) = -15x^2 + 300x
\]
So the equation becomes:
\[
-15x^2 + 300x = 1100
\]
5. Rearranging this gives us the quadratic equation:
\[
-15x^2 + 300x - 1100 = 0
\]
We can multiply through by -1 to make it standard form:
\[
15x^2 - 300x + 1100 = 0
\]
Now, to find the lowest price Ernesto can charge, we can solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 15\), \(b = -300\), and \(c = 1100\).
6. Calculate \(b^2 - 4ac\):
\[
b^2 - 4ac = (-300)^2 - 4 \cdot 15 \cdot 1100
\]
\[
= 90000 - 66000 = 24000
\]
7. Now apply the quadratic formula:
\[
x = \frac{-(-300) \pm \sqrt{24000}}{2 \cdot 15}
\]
\[
= \frac{300 \pm \sqrt{24000}}{30}
\]
8. Calculate \(\sqrt{24000}\):
\[
\sqrt{24000} \approx 155
\]
(rounding for simplicity)
9. Now substitute back into the formula:
\[
x = \frac{300 \pm 155}{30}
\]
This gives us two potential solutions:
\[
x_1 = \frac{455}{30} \approx 15.17 \quad \text{and} \quad x_2 = \frac{145}{30} \approx 4.83
\]
10. Since we need the lowest price that still meets the requirement of $1100 in profit, we take the higher value (since as x decreases, profit decreases). So we round \(15.17\) to the nearest dollar.
Thus, the lowest price Ernesto can charge per download to earn $1,100 per month in profit is:
\[
\boxed{15}
\]