Question

Your school sells yearbooks every spring. The total profit p made depends on the amount x
the school charges for each yearbook. The profit is modeled by the equation p= -2x^2 + 70x
+ 520. What is the smallest amount in dollars the school can charge for a yearbook and
make a profit of at least $1000?​

Answer 1

Answer:

x=25.63

Step-by-step explanation:

turn p into 1000.

1000= -2x^2+70x+520

Subtract 520 from both sides

1000-520= 480

480= -2x^2+70x

Apply the quadratic formula, and x=25.63

So you have to charge at least $25.63

Answer 2

Answer:

$9.361 is the smallest amount, approximately.

Step-by-step explanation:

The given expression is

$p=-2x^{2} +70x+520$

Where $p$ is profit and $x$ is money.

Now, with the restriction of making profits of at least $1000, the expression would be

$-2x^{2} +70x+520 \leq 1000$

We need to multiply the inequality by -1/2,

$x^{2} -35x-260 \geq -500$

Now we solve the quadratic expression

$x^{2} -35x-260 +500 \geq 0\\x^{2} -35x+240 \geq 0$

Solving with a calculator, the numbers that satisfy the quadratic expression are approximately $x_{1} \approx 9.361$ and $x_{2} \approx 25.639$

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.