The demand equation is given as:

$q = -6p + 600$

Here, $q$ is the quantity demanded, and $p$ is the price per item.

Step 1: Express the total revenue $R$ as a function of the price $p$.

Total revenue $R$ is given by the product of the quantity demanded $q$ and the price $p$ per item:

$R(p) = p \times q$

Substituting $q = -6p + 600$ into the equation for revenue:

$R(p) = p \times (-6p + 600)$

Expanding the equation:

$R(p) = -6p^2 + 600p$

Thus, the total revenue $R$ as a function of the price $p$ is:

$R(p) = -6p^2 + 600p$

Step 2: Sketch the graph of $R(p) = -6p^2 + 600p$.

This is a quadratic function that opens downwards (since the coefficient of $p^2$ is negative). The graph is a parabola, and we can find its vertex, intercepts, and general shape to sketch it.

a) Find the vertex (maximum point):

The vertex of a quadratic function $ax^2 + bx + c$ occurs at:

$p = -\frac{b}{2a}$

For $R(p) = -6p^2 + 600p$, $a = -6$ and $b = 600$, so:

$p = -\frac{600}{2(-6)} = \frac{600}{12} = 50$

The maximum revenue occurs when $p = 50$. To find the maximum revenue, substitute $p = 50$ into the equation for $R(p)$:

$R(50) = -6(50)^2 + 600(50) = -6(2500) + 30000 = -15000 + 30000 = 15000$

So, the maximum revenue is 15,000 at a price of 50.

b) Find the intercepts:

• p-intercept: Set $R(p) = 0$ and solve for $p$:

$-6p^2 + 600p = 0$

Factor the equation:

$p(-6p + 600) = 0$

So, $p = 0$ or $p = 100$. The price intercepts are $p = 0$ and $p = 100$.

• R-intercept: The $R$-intercept is when $p = 0$, which gives $R(0) = 0$. Therefore, the graph passes through the origin.

Sketch:

The parabola opens downwards with intercepts at $p = 0$ and $p = 100$, and the maximum revenue occurs at $p = 50$ with $R = 15000$.

Would you like me to generate the sketch graph of this function, or do you have any further questions?

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5 Related Questions:

1. What happens to revenue if the price is set higher than 50?
2. How would the demand equation change if the slope was steeper?
3. What is the break-even price where total revenue becomes zero again?
4. How would an increase in demand at every price affect the revenue function?
5. Can we maximize profit instead of revenue using a similar approach?

Tip: The maximum point of a quadratic revenue function represents the optimal price for generating the most revenue.