1.2-bus-models Cost, revenue, and profit

Module 3 — Business and Lines

The big idea How to build R(x) and P(x), and how to find the break-even quantity.

What this is

These are the business-model problems on 1330 W - Bus Models (textbook §1.2). Cost is given. Revenue is price times quantity. Profit is revenue minus the entire cost function. WebAssign randomizes the numbers — use the profit builder below with your coefficients.

Formulas

Cost · already given
C(x)=given in the problemC(x) = \text{given in the problem}

Total cost of making or building x units. Often a quadratic.

Revenue · price × quantity
R(x)=p⋅xR(x) = p \cdot x

p is the selling price per item, or the revenue per thousand square feet. No extra terms. No x².

Profit · revenue minus cost
P(x)=R(x)−C(x)P(x) = R(x) - C(x)

Never cost minus revenue. Subtract every term in C(x).

Method

1. Write revenue

If they sell each unit for p dollars, revenue is only p times x.

R(x)=pxR(x) = px

2. Subtract the whole cost

Distribute the minus through every term of C(x). If C already has a minus on the x² term, subtracting it flips that term to plus.

P(x)=px−(A+Bx+cx2)=−cx2+(p−B)x−AP(x) = px - (A + Bx + cx^{2}) = -cx^{2} + (p-B)x - A

3. Break even to find a profit

Set P(x) = 0 and solve the quadratic. Keep the root inside the allowed domain (pairs 0 to 500, or thousands of square feet 20 to 400). Throw away a huge root outside that window. If the x² coefficient of P is negative, the parabola opens down and profit is positive after the small in-range root. Round up to a whole item if they manufacture pairs or jerseys.

Traps

The sign trap

−(A + Bx + cx²) = −A − Bx − cx². If C(x) = 2.4 + 0.16x − 0.0001x², then −C(x) contributes +0.0001x². Leaving that term negative makes the rest of the problem wrong.

x is not always “the number of square feet”

On the school-construction problem, x is thousands of square feet. A 50,000 sq ft school is x = 50. Cost and profit come out in millions of dollars.

Revenue is not the cost formula with different letters

Do not copy the 0.1x² or the fixed cost A into R(x). Revenue is only price times quantity.

Worked examples

Sample numbers only — not your WebAssign values.

Soccer cleats · sample numbers, not yours

  • Cost. C(x)=3000+8x+0.1x2,0≤x≤500C(x) = 3000 + 8x + 0.1x^{2},\quad 0 \le x \le 500
  • Price. p=120 dollars per pairp = 120\text{ dollars per pair}

Find R(x), P(x), and how many pairs to manufacture to make a profit.

  1. Revenue is price times pairs.
    R(x)=120xR(x) = 120x
  2. Profit is revenue minus the entire cost.
    P(x)=120x−(3000+8x+0.1x2)=−0.1x2+112x−3000P(x) = 120x - (3000 + 8x + 0.1x^{2}) = -0.1x^{2} + 112x - 3000
  3. Set P(x) = 0. Multiply by −10 to clear the decimal, then use the quadratic formula.
    x2−1120x+30000=0x^{2} - 1120x + 30000 = 0
  4. Roots are about 27.7 and 1092. Only 27.7 is in 0 to 500. The parabola opens down, so profit starts after that root.

Manufacture more than 27 pairs. WebAssign usually wants “more than n” — match the blank.

High school construction · sample numbers

  • Cost (millions of dollars). C(x)=2.4+0.16x−0.0001x2,20≤x≤400C(x) = 2.4 + 0.16x - 0.0001x^{2},\quad 20 \le x \le 400
  • Units. x=thousands of square feetx = \text{thousands of square feet}
  • Revenue rate (sample). r=0.20 million per thousand sq ftr = 0.20\text{ million per thousand sq ft}

Find the profit function. (Your r will be different.)

  1. Revenue is the rate times x.
    R(x)=0.20xR(x) = 0.20x
  2. Subtract C(x). The minus in front of −0.0001x² flips to plus.
    P(x)=0.20x−(2.4+0.16x−0.0001x2)=0.0001x2+0.04x−2.4P(x) = 0.20x - (2.4 + 0.16x - 0.0001x^{2}) = 0.0001x^{2} + 0.04x - 2.4

If part 2 asks you to evaluate, convert square feet to thousands first (80,000 sq ft → x = 80) and leave the answer in millions. If it asks for break-even, set this P(x) = 0 and keep the root between 20 and 400; multiply by 1,000 if they want square feet.