Math  /  Algebra

QuestionThe price p (in dollars) and the quantity xx sold of a certain product satisfy the demand equation x=7p+700x=-7 p+700. 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)=7p2+700pR(p)=-7 p^{2}+700 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 {p0p100}\{p \mid 0 \leq p \leq 100\}. (Simplify your answers. Type integers or decimals.) B. The domain is the set of all real numbers. (c) What price p maximizes revenue? p=$p=\$ (Simplify your answer. Type an integer or a decimal.)

Studdy Solution

STEP 1

What is this asking? We're given how much of a product is sold at a certain price and need to figure out the best price to maximize the total money made. Watch out! Don't mix up *price* and *quantity*!
Also, remember revenue is *price* times *quantity*, not just the price itself.

STEP 2

1. Express revenue as a function of price.
2. Find the domain of the revenue function.
3. Find the price that maximizes revenue.

STEP 3

We're given that x=7p+700x = -7p + 700, where xx is the quantity sold and pp is the price.
Revenue, RR, is calculated as price times quantity, so R=xpR = xp.

STEP 4

Let's **substitute** the expression for xx into the revenue formula: R(p)=(7p+700)pR(p) = (-7p + 700) \cdot p.

STEP 5

**Distribute** pp to both terms inside the parentheses: R(p)=7pp+700pR(p) = -7p \cdot p + 700 \cdot p, which simplifies to R(p)=7p2+700pR(p) = -7p^2 + 700p.
So, our revenue function is R(p)=7p2+700pR(p) = -7p^2 + 700p!

STEP 6

The problem tells us to assume revenue is non-negative, meaning R(p)0R(p) \geq 0.
We also know price can't be negative, so p0p \geq 0.

STEP 7

We have R(p)=7p2+700p0R(p) = -7p^2 + 700p \geq 0.
We can **factor out** 7p-7p: 7p(p100)0-7p(p - 100) \geq 0.

STEP 8

This inequality is true when 0p1000 \leq p \leq 100.
So, the **domain** is {p0p100}\{p \mid 0 \leq p \leq 100\}.
This means the price must be between $0\$0 and $100\$100, inclusive.

STEP 9

Our revenue function R(p)=7p2+700pR(p) = -7p^2 + 700p is a parabola that opens downwards (because the coefficient of p2p^2 is negative).
The maximum revenue occurs at the vertex of this parabola.

STEP 10

The *p*-coordinate of the vertex of a parabola in the form ap2+bp+cap^2 + bp + c is given by b2a-\frac{b}{2a}.
In our case, a=7a = -7 and b=700b = 700.

STEP 11

So, the price that maximizes revenue is p=7002(7)=70014=70014=50p = -\frac{700}{2 \cdot (-7)} = -\frac{700}{-14} = \frac{700}{14} = 50.

STEP 12

(a) The revenue function is R(p)=7p2+700pR(p) = -7p^2 + 700p. (b) The domain of R is {p0p100}\{p \mid 0 \leq p \leq 100\}. (c) The price that maximizes revenue is p=$50p = \$50.

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