Answer:
$9.361 is the smallest amount, approximately.
Explanation:
The given expression is
![p=-2x^(2) +70x+520](https://img.qammunity.org/2020/formulas/mathematics/middle-school/crvltwgsu1nln60seusslrmwtnd8n6ymv8.png)
Where
is profit and
is money.
Now, with the restriction of making profits of at least $1000, the expression would be
![-2x^(2) +70x+520 \leq 1000](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dqowzjxl88lg8c08mn5i6iegmrp9c1xwxn.png)
We need to multiply the inequality by -1/2,
![x^(2) -35x-260 \geq -500](https://img.qammunity.org/2020/formulas/mathematics/middle-school/e6pumvk0tx4shay5k0zjduhbib123fvvb3.png)
Now we solve the quadratic expression
![x^(2) -35x-260 +500 \geq 0\\x^(2) -35x+240 \geq 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vd1traqh4qjw0nhkvsqsmtg2yduvzzn1tg.png)
Solving with a calculator, the numbers that satisfy the quadratic expression are approximately
and
![x_(2) \approx 25.639](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wxxhe0s78vkpt0ptudr59s8f67fte8ysjn.png)
The image attached shows the graph solution of this inequality.
Therefore, the smallest solution to make a profit of at least $1000 is $9.361.