the study of finances of Glassford motors has shown that the profit of the company is modeled by p=-2n^2+800n+370000, where p is profit in dollars, and n is the number of trucks sold
a) at what point (how many trucks sold) will demand be too great (0 profit)
b) how many trucks need to be sold to profit $350000
For part b) put $350000 in for p and solve for n.
For part a), use the method you learned to find the vertex of the parabola
p=-2n^2+800n+370000
To find the number of trucks sold when the profit is zero, you need to solve the equation p = 0.
a) Setting p = 0 in the profit equation p = -2n^2 + 800n + 370000:
0 = -2n^2 + 800n + 370000
To solve this quadratic equation, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -2, b = 800, and c = 370000.
x = (-800 ± √(800^2 - 4(-2)(370000))) / (2(-2))
Simplifying the equation further, you get:
x = (-800 ± √(640000 + 2960000)) / (-4)
x = (-800 ± √(3600000)) / (-4)
x = (-800 ± 1900) / (-4)
Now, calculate both solutions:
x₁ = (-800 + 1900) / (-4) = 275
x₂ = (-800 - 1900) / (-4) = 275
Therefore, the point at which the demand is too great (zero profit) is when 275 trucks are sold.
b) To determine the number of trucks needed to make a profit of $350,000, we need to solve the equation p = 350,000.
Substitute p = 350,000 into the profit equation p = -2n^2 + 800n + 370,000:
350,000 = -2n^2 + 800n + 370,000
Now, rearrange the equation to a quadratic form to solve for n:
2n^2 - 800n + 20,000 = 0
Solving this quadratic equation using the quadratic formula once again, with a = 2, b = -800, and c = 20,000:
n = (-(-800) ± √((-800)^2 - 4(2)(20,000))) / (2(2))
n = (800 ± √(640,000 - 160,000)) / 4
n = (800 ± √(480,000)) / 4
n = (800 ± 692.82) / 4
Now, calculate both solutions:
n₁ = (800 + 692.82) / 4 ≈ 373.21
n₂ = (800 - 692.82) / 4 ≈ 26.79
Therefore, approximately 373 trucks need to be sold to make a profit of $350,000.