Here is our profit as a function of # of posters
p(x) =-10x² + 200x - 250
Here is our price per poster, as a function of the # of posters:
pr(x) = 20 - x
Since we want to find the optimum price and # of posters, let's plug our price function into our profit function, to find the optimum x, and then use that to find the optimum price:
p(x) = -10 (20-x)² + 200 (20 - x) - 250
p(x) = -10 (400 -40x + x²) + 4000 - 200x - 250
Take a look at our profit function. It is a normal trinomial square, with a negative sign on the squared term. This means the curve is a downward facing parabola, so our profit maximum will be the top of the curve.
By taking the derivative, we can find where p'(x) = 0 (where the slope of p(x) equals 0), to see where the top of profit function is.
p(x) = -4000 +400x -10x² + 4000 -200x -250
p'(x) = 400 - 20x -200
0 = 200 - 20x
20x = 200
x = 10
p'(x) = 0 at x=10. This is the peak of our profit function. To find the price per poster, plug x=10 into our price function:
price = 20 - x
price = 10
Now plug x=10 into our original profit function in order to find our maximum profit:
<span>p(x)= -10x^2 +200x -250
p(x) = -10 (10)</span>² +200 (10) - 250
<span>p(x) = -1000 + 2000 - 250
p(x) = 750
Correct answer is C)</span>
Answer: 150 boys
Step-by-step explanation:
Let the number of boys be x
Since there were 200 more girls than boys, the number of girls will be: = x + 200
Since the ratio of boys to girls was 3:7, this can be solved further below:
3/7 = x/x+200
Cross multiply
(7 × x) = 3(x + 200)
7x = 3x + 600
7x - 3x = 600
4x = 600
x = 600/4
x = 150
The number of boys enrolled is 150
Number of Girls will be:
= X + 200
= 150 + 200
= 350
Check: Number of boys / Number of girls
= 150/350
= 3/7
= 3:7
Answer:
Step-by-step explanation:
42 ÷ 7 = 6
Answer:
z=30
Step-by-step explanation:
