# Suppose that the revenue generated by selling x units of a certain commodity is given by the equation R=-3X^2+100X. Assume that R is in dollars. What is the maximum revenue possible in this...

Suppose that the revenue generated by selling x units of a certain commodity is given by the equation R=-3X^2+100X. Assume that R is in dollars. What is the maximum revenue possible in this situation? How many units must be sold to maximize the revenue?

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### 2 Answers

The given revenue function is a quadratic with negative leading coefficient. So its graph is a parabola that is concave down. That means that its vertex is the maximum point of the function.

To get the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, apply the formula

`h = -b/(2a)`

and

`k = f(h)`

In our function

`R = -3x^2 + 100x`

the values of a and b are -3 and 100, respectively. Plugging them to the formula above, would result to:

`h=-100/(2*(-3))=-100/(-6) = 16.67`

`k = -3(16.67)^2+100(16.67) = 833.33`

So, the vertex of the function is (16.67,833.33).

**Therefore, the maximum revenue possible is $833.33 .**

For the number of units to be sold in order to maximize the revenue, refer to the x-coordinate of the vertex. The x-coordinate is 16.67. Since it is in decimal form, we have to round up.

**Therefore, 17 units must be sold to maximize the revenue.**

*(Note: Attach is the graph of the revenue function. )*

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