Question

Find the quantity q that maximizes profit if the total revenue, R(q), and total cost, C(q) are given in dollars by R(q)=3q−0.002q2 and C(q)=200+2.2q, where 0≤q≤1500 units

Answer 1

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

$R(q)=3q-0.002q^{2}$ ---> the revenue

$C(q)=200+2.2q$ ---> the cost

The profit  P(q) is equal to

$P(q)=R(q)-C(q)$

substitute the given values

$P(q)=(3q-0.002q^{2})-(200+2.2q)$

$P(q)=3q-0.002q^{2}-200-2.2q$

$P(q)=-0.002q^{2}+0.8q-200$

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