Answer:
The value of q that maximize the profit is q=200 units
Step-by-step explanation:
we know that
The profit is equal to the revenue minus the cost
we have
[tex]R(q)=3q-0.002q^{2}[/tex] ---> the revenue
[tex]C(q)=200+2.2q[/tex] ---> the cost
The profit P(q) is equal to
[tex]P(q)=R(q)-C(q)[/tex]
substitute the given values
[tex]P(q)=(3q-0.002q^{2})-(200+2.2q)[/tex]
[tex]P(q)=3q-0.002q^{2}-200-2.2q[/tex]
[tex]P(q)=-0.002q^{2}+0.8q-200[/tex]
This is a vertical parabola open downward (because the leading coefficient is negative)
The vertex represent a maximum
The x-coordinate of the vertex represent the value of q that maximize the profit
The y-coordinate of the vertex represent the maximum profit
using a graphing tool
Graph the quadratic equation
The vertex is the point (200,-120)
see the attached figure
therefore
The value of q that maximize the profit is q=200 units
Using the vertex of a quadratic equation, it is found that a quantity of 200 items maximizes the profit.
The revenue function is:
[tex]R(q) = 3q - 0.002q^2[/tex]
The cost function is:
[tex]C(q) = 200 + 2.2q[/tex]
Profit is revenue subtracted by cost, thus:
[tex]P(q) = R(q) - C(q)[/tex]
[tex]P(q) = 3q - 0.002q^2 - 200 - 2.2q[/tex]
[tex]P(q) = -0.002q^2 + 0.8q - 200[/tex]
Which is a quadratic equation with [tex]a = -0.002, b = 0.8, c = -200[/tex].
The maximum value happens at the q-value of the vertex, thus:
[tex]q_V = -\frac{b}{2a}[/tex]
[tex]q_V = -\frac{0.8}{2(-0.002)}[/tex]
[tex]q_V = 200[/tex]
The quantity that maximizes profit is of 200 items.
A similar problem is given at https://brainly.com/question/17616316
