Math  /  Algebra

QuestionThe price pp (in dollars) and the quantity xx sold of a certain product satisfy the demand equation x=6p+300x=-6 p+300. Answer parts (a) through ( g ). (a) Find a model that expresses the revenue R as a function of p . (Remember, R=xp\mathrm{R}=\mathrm{xp}.) R(p)=6p2+300pR(p)=-6 p^{2}+300 p (Simplify your answer. Use integers or decimals for any numbers in the expression.) (b) What is the domain of R ? Assume that R is nonnegative. A. The domain is {p\{p \mid \square p\leq \mathrm{p} \leq \square (Simplify your answers. Type integers or decimals.) B. The domain is the set of all real numbers.

Studdy Solution

STEP 1

What is this asking? We're given how many items (xx) are sold at a given price (pp), and we need to figure out the total revenue (RR) as a function of price, and also what prices make sense in this situation. Watch out! Remember revenue is the total money made, not the profit!
Also, we can't sell negative items, and the price can't be negative either!

STEP 2

1. Express Revenue as a Function of Price
2. Determine the Domain of the Revenue Function

STEP 3

Alright, so we know that revenue is the total money made from selling a certain number of items at a given price.
Mathematically, we can express this as R=xpR = x \cdot p, where RR is the **revenue**, xx is the **quantity sold**, and pp is the **price** per item.

STEP 4

We're given the demand equation x=6p+300x = -6p + 300, which tells us how many items (xx) are sold at a given price (pp).
Let's **substitute** this into our revenue equation: R=(6p+300)pR = (-6p + 300) \cdot p.

STEP 5

Now, let's **distribute** pp to both terms inside the parentheses: R(p)=6p2+300pR(p) = -6p^2 + 300p.
This equation tells us the revenue (RR) as a function of the price (pp).

STEP 6

The problem states that revenue (RR) is non-negative, meaning R(p)0R(p) \geq 0.
So, we have 6p2+300p0-6p^2 + 300p \geq 0.

STEP 7

To find the **domain**, we need to solve this inequality for pp.
First, let's **factor** out 6p-6p: 6p(p50)0-6p(p - 50) \geq 0.
Now, we can **divide** both sides by 6-6 to simplify, remembering to **flip the inequality sign**: p(p50)0p(p - 50) \leq 0.

STEP 8

This inequality holds true when pp and (p50)(p - 50) have opposite signs (or one of them is zero).
This happens when 0p500 \leq p \leq 50.

STEP 9

We know the price (pp) can't be negative.
Also, the quantity sold (xx) must be non-negative, so 6p+3000-6p + 300 \geq 0.
Solving for pp, we get 3006p300 \geq 6p, which simplifies to p50p \leq 50.

STEP 10

Combining the constraints 0p500 \leq p \leq 50 and p50p \leq 50, we find the **domain** of RR is 0p500 \leq p \leq 50.

STEP 11

(a) The revenue function is R(p)=6p2+300pR(p) = -6p^2 + 300p. (b) The domain of RR is {p0p50}\{p \mid 0 \leq p \leq 50 \}.

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