Math Problem Statement

Help me solve the equations

Help me solve the equations.

Solution

To solve the questions in the image, let’s break it down step by step.

Part (a): Complete the Table

The problem gives us:

  • The base admission price is p=10p = 10, with N(p)=25000N(p) = 25000 attendees.
  • For every $1 increase in the admission price, the park loses 250 attendees. This means: N(p)=25000250(p10)N(p) = 25000 - 250(p - 10)
  • The revenue is calculated as R(p)=pN(p)R(p) = p \cdot N(p).

Step 1: Complete the missing values for N(p)N(p) and R(p)R(p)

For p=11p = 11: N(11)=25000250(1110)=250002501=24750N(11) = 25000 - 250(11 - 10) = 25000 - 250 \cdot 1 = 24750 R(11)=1124750=272250R(11) = 11 \cdot 24750 = 272250

For p=12p = 12: N(12)=25000250(1210)=250002502=24500N(12) = 25000 - 250(12 - 10) = 25000 - 250 \cdot 2 = 24500 R(12)=1224500=294000R(12) = 12 \cdot 24500 = 294000

The completed table is:

Admission Price (pp)Attendance (N(p)N(p))Revenue (R(p)R(p))
$1025000$250,000
$1124750$272,250
$1224500$294,000

Part (b): Write the formulas

Formula for N(p)N(p):

As derived earlier, the formula for attendance is: N(p)=25000250(p10)N(p) = 25000 - 250(p - 10) Simplify: N(p)=25000250p+2500N(p) = 25000 - 250p + 2500 N(p)=27500250pN(p) = 27500 - 250p

Formula for R(p)R(p):

The revenue is the price times the number of attendees: R(p)=pN(p)R(p) = p \cdot N(p) Substitute N(p)N(p) into this: R(p)=p(27500250p)R(p) = p \cdot (27500 - 250p) Simplify: R(p)=27500p250p2R(p) = 27500p - 250p^2

So the revenue formula is: R(p)=250p2+27500pR(p) = -250p^2 + 27500p


Part (c): Maximum Revenue

To find the price pp that maximizes revenue, we use the vertex formula for a quadratic equation. The formula for the vertex (maximum point) is: p=b2ap = -\frac{b}{2a} Here, a=250a = -250 and b=27500b = 27500: p=275002(250)=27500500=55p = -\frac{27500}{2(-250)} = \frac{27500}{500} = 55

Thus, the admission price that produces the maximum revenue is p=55p = 55.

Maximum Revenue:

Substitute p=55p = 55 into R(p)R(p): R(55)=250(55)2+27500(55)R(55) = -250(55)^2 + 27500(55) R(55)=250(3025)+27500(55)R(55) = -250(3025) + 27500(55) R(55)=756250+1512500=756250R(55) = -756250 + 1512500 = 756250

The maximum revenue is R(55)=756250R(55) = 756250, and the corresponding attendance is: N(55)=27500250(55)=2750013750=13750N(55) = 27500 - 250(55) = 27500 - 13750 = 13750


Final Results

  1. Completed Table:

    Admission Price (pp)Attendance (N(p)N(p))Revenue (R(p)R(p))
    $1025000$250,000
    $1124750$272,250
    $1224500$294,000
  2. Attendance formula: N(p)=27500250pN(p) = 27500 - 250p

  3. Revenue formula: R(p)=250p2+27500pR(p) = -250p^2 + 27500p

  4. Maximum revenue occurs at p=55p = 55, with R(55)=756250R(55) = 756250 and N(55)=13750N(55) = 13750.

Let me know if you need further clarification or calculations! Here are five related questions for practice:

  1. How does the attendance change when the price increases by $1?
  2. What is the attendance when the price is $30?
  3. What is the revenue when the price is $20?
  4. Can the park achieve revenue higher than $800,000? Why or why not?
  5. Why is the revenue a quadratic function of the price?

Tip: Always check for the domain of your price values when working with quadratic revenue equations!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Relationships
Quadratic Functions
Revenue Maximization

Formulas

N(p) = 27500 - 250p
R(p) = -250p^2 + 27500p
Vertex formula for quadratic functions: p = -b / 2a

Theorems

Vertex theorem for maximum or minimum of a quadratic function

Suitable Grade Level

Grades 9-12