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

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

---> the revenue

---> the cost

The profit  P(q) is equal to

substitute the given values

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:

The cost function is:

Profit is revenue subtracted by cost, thus:

Which is a quadratic equation with .

The maximum value happens at the q-value of the vertex, thus:

The quantity that maximizes profit is of 200 items.

A similar problem is given at link

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24

look at the picture.

y-intercept for the function on a picture is 4 → point (0, 4).

y-intercept for the function

is for x = 0. substitute:

y-intercept is 8 → point (0, 8).